HC Verma Concepts Of Physics Exercise Solutions Of Chapter 3 (Rest and Motion:Kinematics)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 3 - Rest and Motion: Kinematic:

EXERCISE

1. A man has to go 50 m due north, 40 m due east and 20 m due south to reach a field, (a) what distance he has to walk to reach the field? (b) What is his displacement from his house to the field?

Sol:

                       

(a) The distance he has to walk to reach the field is 
                 = (50 + 40 + 20) m =110 m

(b) Magnitude of displacement vector is given by AD

AD = √ {(50 - 20)² + 40²} = √ {900 + 1600} = 50 m and

From ∆ ADE, tan θ = DE/AE = 30/40 = 3/4
                          ⇒ θ = tan^-1 (3/4)

∴ Displacement = 50 m, tan^-1 (3/4) north to east.

2. A particle starts from the origin, goes along the Z-axis to the point (20 m, 0) and then returns along the same line to the point (-20 m, 0). Find the distance and displacement of the particle during the trip. 

 Sol:

                 

Distance travelled by the particle during the trip is

= AB + BC = (20 + 40) m = 60 m

Displacement ➡ shortest distance between final and initial position.

Displacement of the particle during the trip is 
            = 20 m along –ve X direction.

3. It is 260 km from Patna to Ranchi by air and 320 km by road. An aeroplane takes 30 minutes to go from Patna to Ranchi whereas a delux bus takes 8 hours, (a) Find the average speed of the plane, (b) Find the average speed of the bus. (c) Find the average velocity of the plane, (d) Find the average velocity of the bus. 

Sol:

(a) Given: distance from Patna to Ranchi by air = 260 km and time taken by an aeroplane = 30 min = 0.5 hour.

ஃ the average speed of the plane = distance/time
              = 260/0.5 km h^-1 = 520 km h^-1  

(b) Given: distance from Patna to Ranchi by road = 320 km and time taken by delux bus = 8 hours.

ஃ the average speed of the Bus = distance/time 
             = 320/8 km h^-1 = 40 km h^-1  

(c) Displacement ➡ shortest distance between final and initial position

ஃThe average velocity of the plane = displacement/time 
             = 260/0.5 km h^-1 = 520 km h^-1 Patna to Ranchi

(d) Displacement ➡ shortest distance between final and initial position

ஃThe average velocity of the plane = displacement/time 
             = 260/8 km h^-1 = 32.5 km h^-1 Patna to Ranchi

4. When a person leaves his home for sightseeing by his car, the meter reads 12352 km. When he returns home after two hours the reading is 12416 km. (a) what is the average speed of the car during this period? (b) What is the average velocity?

Sol:

(a) The average speed of the car during this period is

= (12416 - 12352)/2 km h^-1 = 32 km h^-1

(b) The average velocity is = displacement/time

Displacement ➡ shortest distance between final and initial position

Displacement = Zero [because initial and final position is same]

ஃThe average velocity is zero.

5. An athelete takes 2.0 s to reach his maximum speed of 18.0 km/h. What is the magnitude of his average acceleration?

Sol:
Maximum speed (v) = 18.0 km/h 
                 = (18 * 1000)/3600 = 5.0 m s^-1.

Athelete start from zero speed, 
        So initial speed (u) = 0 m s^-1.

Time taken (t) = 2 s.

ஃ the magnitude of his average acceleration is

|A _ave| = (v - u)/t = (5 – 0)/2 = 2.5 m s^-2

6. The speed of a car as a function of time is shown in figure (3-E1). Find the distance travelled by the car in 8 seconds and its acceleration.

      

Sol:  

From the graph we get following data:

Initial velocity (u) = 0 m s^-1; final velocity (v) = 20 m s^-1; 
time taken = 8 s.

We know, a = (v - u)/t 
Or, a = (20 - 0)/8 = 2.5 m s^-2

Acceleration, a = 2.5 m s^-2   

Distance, s = ut + ½ (at²) 
Or, s = 0 * 8 + ½ * 2.5 * 8² = 80 m

7. The acceleration of a cart started at t = 0, varies with time as shown in figure (3-E2). Find the distance travelled in 30 seconds and draw the position-time graph.
                         

Sol:

At t = 0, cart started. So initial speed (u) = 0; for 1^st 10 s acceleration, a = 5.0 ft s^-2.

Distance travelled in 1^st 10 s is

S₁ = ut + ½ at² 
    = 0 * 10 + ½ * 5 * 10² = 250 ft.

At 10 s velocity of cart, v_{10 s} = u + at = 0 + 5 * 10 = 50 ft/s; a_10-20 = 0

Distance travelled in 2^nd 10 s is

S₂ = v_{10 s} * t = 50 * 10 = 500 ft/s

For last 10 s: a = - 5.0 ft s^-2; u = 50 ft/s;

Distance travelled in 3^rd 10 s is

S₃ = ut + ½ at² 
    = 50 * 10 + ½ * (-5) * 10² 
    = 500 -250 = 250 ft

Total distance travelled,

           S = S₁ + S₂ + S₃  

              = (250 + 500 +250) ft = 1000 ft

The position-time graph 

                             

8. Figure (3-E3) shows the graph of velocity versus time for a particle going along the X-axis. Find (a) the acceleration, (b) the distance travelled in 0 to 10 s and (c) the displacement in 0 to 10 s.                                                        

Sol:

From the graph we get,

Initial velocity (u) = 2 m/s; final velocity (v) = 8 m/s; 
time taken = 10 s.

(a) acceleration, a = (v - u)/t = (8 - 2)/10 = 0.6 m/s² along +ve x axis.

(b) The distance travelled in 0 to 10 s is given by

S = ut + ½ at² 
   = 2 * 10 + ½ * 0.6 * 10² = 50 m

(c) Displacement = 50 m along +ve x axis.          

9. Figure (3-E4) shows the graph of the x-coordinate of a particle going along the X-axis as a function of time. Find (a) the average velocity during 0 to 10 s, (b) instantaneous velocity at 2, 5, 8 and 12 s.

                       

Sol:

(a) The average velocity during 0 to 10 s is given by, V_ave = displacement/time.

Displacement during 0 to 10 s = 100 m.

V_ave = (100/10) m/s = 10 m/s.

(b) Slope of the curve in position-time graph in each point give us instantaneous velocity.

Velocity is uniform in time interval from 0 to 2.5 s (slope of straight line is constant). So velocity is same for all instant between 0 to 2.5 s.

ஃ Instantaneous velocity at 2 s = tan θ = 50/2.5 = 20 m/s.

From 2.5 s to 7.5 s position of particle is not change. So, particle is at rest in this interval.

ஃ Instantaneous velocity at 5 s = 0 m/s.

From 7.5 s to 10.0 s velocity is uniform.

ஃ Instantaneous velocity at 8 s = (100 - 50)/ (10.0 – 7.5) = 20 m/s

From 10.0 s to 15.0 s velocity is uniform.

ஃ Instantaneous velocity at 8 s = (0 - 100)/ (15.0 – 10.0) = -20 m/s

10. From the velocity-time plot shown in figure (3-E5), find the distance travelled by the particle during the first 40 seconds. Also find the average velocity during this period.

Sol:

The distance travelled by the particle is equal to the area under curve in velocity-time graph.

ஃ the distance travelled by the particle during the first 40 seconds, S = |area (OAB)| + |area (BCD)|
Or, S = ½ * 5 * 20 + ½ * 5 * (40 - 20) = 100 m

Displacement = area (OAB) + area (BCD)

      = ½ * 5 * 20 - ½ * 5 * (40 - 20) = 0 m

Therefore, velocity = 0 m/s

11. Figure (3-E6) shows x-t graph of a particle. Find the time t such that the average velocity of the particle during the period 0 to t is zero.

                     
Sol:

The average velocity of the particle during the period 0 to t is zero, if position at t = t is same as position at t = 0. That means, position S_{t = 0} = 20 m = S_{t = t}.                                                        

From the graph, we can found t = 12 s

12. A particle starts from a point A and travels along the solid curve shown in figure (3-E7). Find approximately the position B of the particle such that the average velocity between the positions A and B has the same direction as the instantaneous velocity at B.

                        

Sol:
                      
Average velocity between the positions A and B is

v_ab = (AB)/time is along AB.

At position B instantaneous velocity has also directed along AB.

From the graph we get, point B is (5 m, 3 m).

13. An object having a velocity 4.0 m/s is accelerated at the rate of 1.2 m/s² for 5.0 s. Find the distance travelled during the period of acceleration.

Sol:
Given: initial velocity (u) = 4.0 m/s; acceleration (a) = 1.2 m/s²; time period (t) = 5.0 s.

We know, distance, S = ut + ½ at²

Or, S = 4 * 5 + ½ * 1.2 * 5² = 35 m.

14. A person travelling at 43.2 km/h applies the brake giving a deceleration of 6.0 m/s² to his scooter. How far will it travel before stopping?

Sol:

Given: u = 43.2 km/h = (43.2 * 1000)/3600 = 12 m/s; final velocity (v) = 0; a = - 6.0 m/s²; S =?

We know, v² – u² = 2*a*s 
Or, s = (v² – u²)/2a

Or, s = (0² – 12²) / {2 *(-6)} = 12 m

15. A train starts from rest and moves with a constant acceleration of 2.0 m/s² for half a minute. The brakes are then applied and the train comes to rest in one minute. Find (a) the total distance moved by the train, (b) the maximum speed attained by the train and (c) the position(s) of the train at half the maximum speed.

Sol:

Train is starting from rest and moves with constant acceleration and attain the maximum speed. After that it start decelerating and it comes to rest. Velocity-time graph shown below.

From 0 to 30s, we have

u = 0; a = 2.0 m/s²; t = 30 s.

Final velocity, v is given by v = u + at

ஃ v = 0 + 2 * 30 = 60 m/s and

Distance travelled from 0 to 30 s is,

          s₁ = ut + ½ at²

              = 0 * 30 + ½ * 2 * 30² = 900 m

From 30 s to 90 s, we have

v = 0; u = final velocity of 1^st time period = 60 m/s; t = 60 s;

Acceleration, a = (v - u)/t = - 1 m/s,

Distance travelled from 30 to 90 s is,

s₂ = (v² – u²)/2a = (0² – 60²)/2(-1) = 1800 m

(a) The total distance moved by the train is

S = s₁ + s₂ = 900 m + 1800 m = 2700 m = 2.7 km

(b) The maximum speed attained by the train, v_max = 60 m/s.

(c) Half the maximum speed = 60/2= 30 m/s

Positions of the train at v_max/2 are

Two times train attain half of max. Velocity

(i) In the time of acceleration: position of train from starting is given by,

S = (v² – u²)/2a = (30² – 0²)/ (2*2) = 225 m

(ii) In the time of deceleration: position of train from maximum velocity is given by,

S = (v² – u²)/2a = (30² – 60²)/2(-1) = 1350 m

Position of train from staring = 900 m + 1350 m

                             = 2250 m =2.25 km

16. A bullet travelling with a velocity of 16 m/s penetrates a tree trunk and comes to rest in 0.4 m. Find the time taken during the retardation.

Sol:

Given: u = 16 m/s; v = 0; s = 0.4 m; t =?

First of all we have to find the retardation (a), which is given by

2as = v² + u² → a = (v² – u²)/2s 
Or, a = (0² – 16²)/ (2*0.4) = - 320 m/s²

Now, we know

→ a * t = v – u or, t = (v - u)/ a

Or, t = (0 – 16)/ (- 320) = 0.05 s

17. A bullet going with speed 350 m/s enters a concrete wall and penetrates a distance of 5.0 cm before coming at rest. Find the deceleration.

Sol:

Given: u = 350 m/s; v = 0 m/s; s = 5.0 cm = 0.05 m; a =?
We know, 2as = v² – u²

Or, a = (v² – u²)/2s = (0² – 350²)/ (2*0.05)
         = - 12.25 * 10⁵ m/s²  

18. A particle starting from rest moves with constant acceleration. If it takes 5.0 s to reach the speed 18.0 km/h find (a) the average velocity during this period, and (b) the distance travelled by the particle during this period.

Sol:

Given: u = 0 m/s; v = 18.0 km/h = 5 m/s; t = 5.0 s;

Acceleration, a = (v – u)/t = (5 – 0)/5 = 1 m/s²;

Distance travelled, s = ut + ½ at² = 0*5 + ½ * 1* 5² = 12.5 m

 (a) The average velocity during this period,

V_ave = (final position – initial position)/time 
        = (12.5 - 0)/5 = 2.5 m/s 

(b) The distance travelled by the particle during this period is 12.5 m.

19. A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time of this driver. If he is driving a car at a speed of 54 km/h and the brakes cause a deceleration of 6.0 m/s², find the distance travelled by the car after he sees the need to put the brakes on.

Sol:
Given: u = 54 km/h = 15 m/s; v = 0 m/s; a = - 6.0 m/s²; reaction time = 0.20 s.

From 0 to reaction time car will move with uniform velocity (15 m/s).

Therefore, car moves in this period of time is s₁ = 15 * 0.2 = 3 m.

Distance travelled in the time of deceleration is given by,

S₂ = (v² – u²)/2a = (0² – 15²)/2(- 6) = 18.75 m

Therefore, the distance travelled by the car after he sees the need to put the brakes on is

S = s₁ + s₂ = (3 + 18.75) m = 21.75 m ≈ 22 m

20. Complete the following table:

Sol:

For car A and driver X:

Given: u = 54 km/h = 15 m/s; v = 0 m/s; a = - 6.0 m/s²; reaction time = 0.25 s.

From 0 to reaction time car will move with uniform velocity (15 m/s).

Therefore, car moves in this period of time is s₁ = 15 * 0.25 = 3.75 m.

Distance travelled in the time of deceleration is given by,

S₂ = (v² – u²)/2a = (0² – 15²)/2(- 6) = 18.75 m

Therefore, braking distance, a = 19 m

Total stopping distance, b = s₁ + s₂ 
                    = (3.75 + 18.75) m = 22.5 m

For car A and driver Y:

Given: u = 72 km/h = 20 m/s; v = 0 m/s; a = - 6.0 m/s²; reaction time = 0.3 s.

From 0 to reaction time car will move with uniform velocity (20 m/s).

Therefore, car moves in this period of time is s₁ = 20 * 0.3 = 6 m.

Distance travelled in the time of deceleration is given by,

S₂ = (v² – u²)/2a = (0² – 20²)/2(- 6) = 33.3 m
Therefore, braking distance, c = 33 m

Total stopping distance, d = s₁ + s₂ 
                           = (6 + 33) m = 39 m

For car B and driver X:

Given: u = 54 km/h = 15 m/s; v = 0 m/s; a = - 7.5 m/s²; reaction time = 0.20 s.

From 0 to reaction time car will move with uniform velocity (15 m/s).

Therefore, car moves in this period of time is s₁ = 15 * 0.20 = 3 m.

Distance travelled in the time of deceleration is given by,

S₂ = (v² – u²)/2a = (0² – 15²)/2(- 7.5) = 15 m

Therefore, braking distance, e = 15 m

Total stopping distance, f = s₁ + s₂ 
                             = (3 + 15) m = 18 m

For car B and driver Y:

Given: u = 72 km/h = 20 m/s; v = 0 m/s; a = - 7.5 m/s²; 
reaction time = 0.3 s.

From 0 to reaction time car will move with uniform velocity (20 m/s).

Therefore, car moves in this period of time is s₁ = 20 * 0.3 = 6 m.

Distance travelled in the time of deceleration is given by,

S₂ = (v² – u²)/2a = (0² – 20²)/2(- 7.5) = 26.6 m

Therefore, braking distance, g = 27 m

Total stopping distance, h = s₁ + s₂ 
                            = (6 + 27) m = 33 m

21. A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of 72 km/h. The jeep follows it at a speed of 90 km/h, crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike?

Sol:

Given: Speed of the jeep, v_j = 90 km/h = 25 m/s; speed of bike, v_b = 72 km/h = 20 m/s; jeep cross the turning 10 s later than the bike.

Bike is (20 * 10) = 200 m ahead from the jeep, when jeep crossing the turning.

             

Let, we assume that the jeep will catch up with the bike at C.

From this instant, both jeep and bike takes same time to reach at C.

Therefore, 

          time taken by jeep (t_j) = time taken by bike (t_b)

Or, (dist. travelled, S_b)/v_j = (dist. travelled, S_b)/v_b  

Or, (200 + x)/25 = x/20 ⇒ x = 800 m.

Total distance travelled by jeep from the turning is (200 + 800) m = 1000 m =1.0 km

22. A car travelling at 60 km/h overtakes another car travelling at 42 km/h. Assuming each car to be 5.0 m long, find the time taken during the overtake and the total road distance used for the overtake.

Sol:     

Speed of the car A is 60 km/h = 50/3 m/s and speed of car B is 42 km/h = 35/3 m/s.

Distance travelled by car A during the time of overtake is

                 = (5 + x + 5 + 5) m = (x +15) m.

During this time distance travelled by car B is = (x + 5) m

Both take the same time to cover their distance.

Therefore, (x +15)/ (50/3) = (x + 5)/ (35/3)

Or, 7 x + 105 = 10 x + 50 ⇒ x = 55/3 m

So, the total distance travelled by car A is

= (55/3 +15) m = 100/3

Therefore, the time taken during the overtake is

= (100/3)/ (50/3) = 2 s.

And the total road distance used for the overtake is

= distance travelled by car A + length of the car

= 100/3 + 5 = 38.3 m.

23. A ball is projected vertically upward with a speed of 50 m/s. Find (a) the maximum height, (b) the time to reach the maximum height, (c) the speed at half the maximum height. Take g = 10 m/s².

Sol:
Given: u = 50 m/s; θ = 90⁰; g = - 10 m/s²;

(a) At maximum height final velocity, v = 0 m/s;

We know, v² - u² = 2*a*s

Or, 0² - 50² = 2 * (- 10) * H

Or, H = 2500/20 = 125 m

(b) We know, v – u = a * t

Or, 0 – 50 = - 10 * t ⇒ t = 50/10 = 5 s 

The time to reach the maximum height is 5 s.

(c) S = 125/2 m

We know, v² - u² = 2 * a * s

Or, v² - 50² = 2* (-10) * (125/2)

Or, v² = 2500 – 1250 = 1250

Or, v = √ (1250) = 35.36 m/s ⋍ 35 m/s

24. A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height 60 m at the time of dropping the ball, how long will the ball take in reaching the ground?

Sol:
Given: h = - 60 m; h = 7 m/s; g = - 10 m/s

We know, s = u*t + ½ * a* t²

Or, - 60 = 7 * t + ½ * (- 10) * t²

Or, 5 t² - 7 t - 60 = 0     

Or, t = [- (- 7) ± √ {(- 7)² – 4*5*(-60)}]/ {2*5}  

Or, t = {7 ± (√1249)}/10 = - 2.8 and 4.3

Time cannot be –ve, therefore t is = 4.3 s

25. A stone is thrown vertically upward with a speed of 28 m/s. (a) Find the maximum height reached by the stone, (b) Find its velocity one second before it reaches the maximum height, (c) Does the answer of part (b) change if the initial speed is more than 28 m/s such as 40 m/s or 80 m/s?

Sol:

Given: u = 28 m/s; g = - 9.8 m/s

(a) At the maximum height, final velocity (v) = 0 m/s.

We know, v² – u² = 2 * a * s

Or, 0² – 28² = 2* (- 9.8)* S; or, S = 40 m

Therefore, the maximum height reached by the stone is 40 m.

(b) Time taken to reach maximum height is given by

 t = (v – u)/a

Or, t = (0 – 28)/ (- 9.8) = 2.85 s

For the given question new time, t’ = 2.85 – 1 = 1.85 s

Velocity one second before it reaches the maximum height is given by

V = u + at = 28 + (- 9.8)*1.85 = 9.8 m/s.

(c) If u = 40 m/s for question (b).

Time taken to reach maximum height is given by 

t = (v – u)/a

Or, t = (0 – 40)/ (- 9.8) = 4.08 s

For the given question new time, t’ = 4.08 – 1 = 3.08 s

Velocity one second before it reaches the maximum height is given by

V = u + at = 40 + (- 9.8)*3.08 = 9.8 m/s.

Therefore, the answer of part (b) not change if the initial speed is more than 28 m/s such as 40 m/s or 80 m/s.

26. A person sitting on the top of a tall building is dropping balls at regular intervals of one second. Find the positions of the 3rd, 4th and 5th ball when the 6th ball is being dropped.

Sol:

Initial velocity (u) = 0 m/s; g = - 9.8 m/s²;

Time of flight for 3^rd ball when the 6th ball is being dropped is t = 3 s.

Distance travelled by 3^rd ball is

S = u*t + ½ at² 

  = 0 * 3 + ½ *(- 9.8) * 3² = - 44.1 m

The positions of the 3^rd is 44.1 m below from the top.

Time of flight for 3^rd ball when the 6th ball is being dropped is given by t = 3 s.

Time of flight for 4^th ball when the 6th ball is being dropped is given by t = 2 s.

Distance travelled by 4^th ball is

S = u*t + ½ at² = 0 * 2 + ½ *(- 9.8) * 2² = - 19.6 m

The positions of the 4^th ball is 19.6 m below from the top.

Time of flight for 5^th ball when the 6th ball is being dropped is given by t = 1 s.

Distance travelled by 5^th ball is

S = u*t + ½ at² = 0 * 1 + ½ *(- 9.8) * 1² = - 4.9 m

The positions of the 5^th ball is 4.9 m below from the top.

27. A healthy youngman standing at a distance of 7 m from a 11.8 m high building sees a kid slipping from the top floor. With what speed (assumed uniform) should he run to catch the kid at the arms height (1.8 m)?

Sol:

We know, s = u * t + ½ at²

Or, - 10 = 0 * t + ½ * (- 9.8) * t²

Or, t = 1.4 s

The kid takes time to reach the ground is 
1.4 s.

Distance travelled by the man to catch the kid is = 7 m.

In this time the man has to reach at the bottom of the building.

Therefore, velocity the man is = s/t = 7/1.42 = 4.9 m/s.

28. An NCC parade is going at a uniform speed of 6 km/h through a place under a berry tree on which a bird is sitting at a height of 12.1 m. At a particular instant the bird drops a berry. Which cadet (give the distance from the tree at the instant) will receive the berry on his uniform?
Sol:

Let the distance of the cadet from the tree at the instant be x.

Speed of the cadet, v = 6 km/h = 1.67 m/s.

Initial speed of berry fruit = 0 m/s.

The time taken by berry to reach the ground is given by

         S = u*t + ½ * a* t²

Or, - 12.1 = 0*t + ½ * (- 9.8) * t²

Or, t = 1.57 s

X = v_cadet * time = (1.67 * 1.57) m = 2.6 m

The cadet, 2.6 m away from tree will receive the berry on his uniform.

29. A ball is dropped from a height. If it takes 0.200 s to cross the last 6.00 m before hitting the ground, find the height from which it was dropped. Take g = 10 m/s.

Sol:

Let we assume that height of the building 

be (x + 6) m.

Initial velocity of the ball is u = 0 m/s. Velocity of the ball after (x) m is given by 

                      v² - u² = 2*a*s

Or, v² - 0² = 2* (- 10)* (-x)

Or, v = √ (20 x)

Now, for last 6 m

Initial velocity,  u = √ (20 x); t = 0.200 s; s = 6 m;
We know, s = u * t + ½ * a * t² 

Or, - 6 = - √ (20 x) * 0.2 + ½ * (- 10) * (0.2)² 
Or, √ (20 x) * 0.2 = 6.2, ⇒ x = 48 m

30. A ball is dropped from a height of 5 m onto a sandy floor and penetrates the sand up to 10 cm before coming to rest. Find the retardation of the ball in sand assuming it to be uniform.

Sol:
Given: s = height = - 5 m; u = 0 m/s; g = - 9.8 m/s²;

Velocity of the ball when it touches the sandy floor is given by

                   v² – u² = 2 * a * s

Or, v² – 0² = 2 * (- 9.8) * (- 5)

Or, v = 9.89 m/s

For the motion of ball in sand

We have, u = 9.89 m/s; s = 10 cm = 0.1 m; v = 0 m/s;

We know, a = (v² – u²)/2s

Or, a = (0² – 9.89²)/ (2*0.1) = - 490 m/s².

The retardation of the ball in sand is 490 m/s².

31. An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 6 ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.

Sol:

Given: initial velocity of the coin and elevator are u_c = 0 and u_e = 0; acceleration ball, a_c = - g; t_c = 1 s; t_e = 1 s; g = 32.2 ft/s²

Let we assume that, the acceleration of elevator be a.
In this period of time, coin travelled from c to c’ and elevator travelled from ab to a’b’.

Distance, cc’ = u_c * t_c + ½ *a_c*t_c² 

             = 0 * 1 + ½ * (- 32.2) *1² = - 16.1 ft.
Distance, ab-a’b’ = u_e * t_e + ½ *a_e*t_e² 

                     = 0 * 1 + ½ * (a) *1² = a/2 ft.
According to the question,

         (Distance, cc’) – (Distance, ab-a’b’) = - 6

Or, - 16.1 – (a/2) = - 6

Or, a = - 20.2 ft/s². (-ve sign shows acceleration is downward)

Deceleration of the elevator is 20.2 ft/s².

32. A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s. Find (a) the time it takes to reach the ground, (b) the horizontal distance it travels before reaching the ground, (c) the velocity (direction and magnitude) with which it strikes the ground.

Sol:

Given: at the point A, u_HA = 200 m/s; u_VA = 0; s_V = - 100 m; a_V = - g = - 9.8 m/s²; a_H = 0 therefore horizontal velocity remain constant (u_HA = v_HB)

(a) We know,

          s_V = u_VA * t + ½ *a_V * t²

Or, - 100 = 0 * t + ½ * (- 9.8) * t²

Or, t = 4.5 s

The time it takes to reach the ground is = 4.5 s.

(b) S_H = horizontal speed * time of flight 

           = 20 * 4.5 = 90 m

(c) Vertical velocity at point B is v_VB.

We know, (v_VB)² + (u_VA)² = 2 * a_V * s_V

Or, (v_VB)² + (0)² = 2 * (- 9.8) * (- 100) 

Or, (v_VB)² = 1960 → v_v = 44.2 m/s

Horizontal velocity at point B is, v_HB = u_HA = 20 m/s [∵ a_H = 0]

Therefore, the magnitude of velocity at B is, 

  v = √ {(v_VB)² + (v_HB)²} 

= √ {(44.2)² + (20)²} 
= √ {1920 + 400} = 48.16 m/s.

Angle between resultant and horizontal is θ.

Tan θ = (v_VB)/ (v_HB) = 44.2/20 = 2.21

Or, θ = Tan^-1 2.21 = 65.65⁰ ⋍ 66⁰..

33. A ball is thrown at a speed of 40 m/s at an angle of 60° with the horizontal. Find (a) the maximum height reached and (b) the range of the ball. Take g = 10 m/s.

Sol:

Given: u = 40 m/s; θ = 60°; g = 10 m/s;

(a) We know, maximum height, 

                 H_max = (u² sin² θ)/2g

Or, H_max = (40² sin² 60⁰)/ (2*10) = 60 m.   

(b) We know,

            Range, R = (u² sin 2θ)/g
Or, R = (40² sin 2*60)/10 = (80√3) = 138.56 m.

34. In a soccer practice session the football is kept at the centre of the field 40 yards from the 10 ft high goalposts. A goal is attempted by kicking the football at a speed of 64 ft/s at an angle of 45° to the horizontal. Will the ball reach the goal post?

Sol:

Given: θ =45⁰; u = 64 ft/s; R = 40 yards = (40*3) ft = 120 ft;

Horizontal component of velocity = u cos 45⁰

Time taken to reach the goal post = range/u cos 45⁰

                              = 120/ {64 *(1/√2)} = 2.65 s

Position of ball along y-axis after 2.65 s

→ y = (u*sin 45⁰) t – ½ *g*t²

Or, y = (64*0.707)*2.65 – ½ * 32.2 *(2.65)²

Or, y = 120 – 113 = 7 ft (which is less than height of goalpost)

In time 2.65, the ball travels vertical height 7.0 ft which is less than 10 ft. The ball will reach the goal post.

35. A popular game in Indian villages is goli which is played with small glass balls called golis. The goli of one player is situated at a distance of 2.0 m from the goli of the second player. This second player has to project his goli by keeping the thumb of the left hand at the place of his goli, holding the goli between his two middle fingers and making the throw. If the projected goli hits the goli of the first player, the second player wins. If the height from which the goli is projected is 19.6 cm from the ground and the goli is to be projected horizontally, with what speed should it be projected so that it directly hits the stationary goli without falling on the ground earlier?

Sol:

Given: height of the projection, h = 19.6 cm = 0.196 m; distance between two golis, R = 2 m; horizontal speed, u =?? 

We know, s_v = u_v * t + ½ *a_v* t²

Or, - 0.196 = 0 *t + ½ * (- 9.8)*t²

Or, t = √ (0.196/4.9) = 0.2 s

Therefore, time taken by goli to hit the stationary goli is 0.2 s.

We know,

Range, R = horizontal speed * time of flight

Or, 2 = u * 0.2 → u = 2/0.2 = 10 m/s.

36. Figure (3-E8) shows a 11.7 ft wide ditch with the approach roads at an angle of 15° with the horizontal. With what minimum speed should a motorbike be moving on the road so that it safely crosses the ditch?

Assume that the length of the bike is 5 ft, and it leaves the road when the front part runs out of the approach road.
Sol: 

Given: Horizontal range X = 11.7 + 5 = 16.7 ft covered by the bike.

g = 9.8 m/s² = 32.2 ft/s²; θ = 15⁰

y = x tan θ – ½ *(gx² sec² θ)/ (2u²)

For minimum speed, y = 0

Therefore, x tan θ = ½ *(gx² sec² θ)/ (u²)  

Or, u² = gx/ (sin 2 θ)              

Or, u = √ {gx/ (sin 2 θ)} = 32.8 ft/s  

37. A person standing on the top of a cliff 171 ft high has to throw a packet to his friend standing on the ground 228 ft horizontally away. If he throws the packet directly aiming at the friend with a speed of 15.0 ft/s, how short will the packet fall?

Sol:

Given: height of projection, h = 171 ft; u = 15 ft/s and makes angle θ with horizontal. 

Tan θ = (171)/ (228) → θ = tan^-1 (171/228) = 36.9⁰

Vertical motion:

→ y = u_y*t + ½ * a_y * t²

Or, - 171 = - (15*sin 36.9⁰)*t + ½ *(- 32.2)*t²

Or, 16t² – 9t – 171 = 0

Or, t = 3 and – 3.56

Time cannot be –ve, therefore time of flight, t = 3 s

Horizontal motion:

Range, R = u_x * time of flight 

                = 15 cos 36.9⁰ * 3 = 36 ft

The packet is fall (228 – 36) = 192 ft short. 

38. A ball is projected from a point on the floor with a speed of 15 m/s at an angle of 60° with the horizontal. Will it hit a vertical wall 5 m away from the point of projection and perpendicular to the plane of projection without hitting the floor? Will the answer differ if the wall is 22 m away?

Sol:

Given: u = 15 m/s; θ = 60⁰;

Range, R = (u² sin 2θ)/g 

           = (15² * sin 120⁰)/9.8 = 19.88 m

In first case the wall is 5 m away from projection point, so it is in the horizontal range of projectile. So the ball will hit the wall. In second case (22 m away) wall is not within the horizontal range. So the ball would not hit the wall.

39. Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with a speed u at an angle 6 with the horizontal.

Sol:

Average velocity, V_ave = displacement/time

For the given instant, vertical displacement is zero and horizontal displacement is AB. So there is no effect of vertical component of the velocity during this displacement.

In this instant horizontal velocity remains constant (u cos θ).

Therefore, V_ave = u cos θ along horizontal direction.

40. A bomb is dropped from a plane flying horizontally with uniform speed. Show that the bomb will explode vertically below the plane. Is the statement true if the plane flies with uniform speed but not horizontally?
Sol:

Let we assume the bomb explode i.e. reach the ground in time t. in this instant of time, distance travelled by plane is u*t. Distance travelled in horizontal direction by bomb i.e. Range, R = u *t. Both the distances are same. Therefore, the bomb will explode vertically below the plane.

Suppose the aeroplane move making angle θ with horizontal. For both bomb and aeroplane, horizontal distance is u cos θ * t. t is time for bomb to reach the ground.

So in this case also, the bomb will explode vertically below aeroplane.

41. A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1 m/s² and the projection velocity in the vertical direction is 9.8 m/s. How far behind the boy will the ball fall on the car?

Sol:

Let the velocity of car be u when the ball is thrown. Initial velocity of car is = Horizontal velocity of ball.

Time of flight of the ball, 
t = (2u sin 90⁰)/g = (2*9.8)/9.8 = 2 s.

Distance travelled by car in this period of time is

S_c = u*t + ½ at² = (2u + 2) m.

Range of the ball, S_b = u*t = (2u) m.

S_c - S_b = (2u + 2) - (2u) = 2 m.

Therefore, the ball will drop 2 m behind the boy.

42. A staircase contains three steps each 10 cm high and 20 cm wide (figure 3-E9). What should be the minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane?

Sol:

At minimum velocity it will move just touching point C reaching the ground. So, Range, R is = OC = 0.4 m and Height, H is = OA = - 0.2 m.

Let the horizontal velocity of a ball be u.

Vertical motion: u = 0; a = - g = 9.8 m/s²; s = - 0.2 m

We know, s = u*t + ½ *a* t²

Or, - 0.2 = 0 * t + ½ * (- 9.8) * t²

Or, t = √ (2/49) = 0.2 s

Horizontal motion:

We know, Range, R = horizontal velocity * time of flight

Or, 0.4 = u * 0.2

Or, u = 2 m/s

So, the horizontal velocity of a ball is 2 m/s.


43. A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed and the angle of projection (a) as seen from the truck, (b) as seen from the road.
Sol:
Given: horizontal velocity of truck = u_x =14.7 m/s; distance travelled by truck = horizontal distance travelled by ball = 58.8 m;

Time of flight of the ball = time taken by truck to reach 58.8 m = (58.8)/ (14.7) = 4 s 

Therefore, time taken by ball to reach maximum is = 4/2 = 2 s.

Vertical motion of the ball:

Initial velocity = u; final velocity, v = 0; t = 2 s; a = - g = - 9.8 m/s²;

We have, v - u = at

Or, 0 - u = (- 9.8)* 2

Or, u = 19.6 m/s

(a) The speed and the angle of projection, as seen from the truck is = 19.6 m/s upward.

(b) The speed and the angle of projection, as seen from the road is 

= √ {(u_x)² + (u_y)²} = √ {(14.7)² + (19.6)²} = 24.5 m/s and 

θ = tan^-1 (u_y/u_x) = tan^-1 (19.6/14.7) = 53⁰ with horizontal.

44. The benches of a gallery in a cricket stadium are 1 m wide and 1 m high. A batsman strikes the ball at a level one metre above the ground and hits a mammoth sixer. The ball starts at 35 m/s at an angle of 53° with the horizontal. The benches are perpendicular to the plane of motion and the first bench is 110 m from the batsman. On which bench will the ball hit?

Sol:

Suppose ball hit the n^th bench.

When ball hit the bench:

X- Coordinate = (110 + n – 1) m

Y- Coordinate = (n - 1) m

We know, y = (x tan θ) + ½ * a * {x²/ (u² cos² θ)}

Or, (n - 1) = (110 + n – 1) tan 53⁰ 

                      + ½ *(- 9.8)* {(110 + n – 1)²/ (35² cos² 53⁰)}

Or, (n -1) = (109 + n)*1.327 – 0.011*(109 + n)²

Or, n – 1 = 144.65 + 1.327 n – 0.011 *(n² + 218 n + 11881)

Or, 0.011 n² + 2.071 n – 14.959 = 0

Solving this equation we get, n = 6

45. A man is sitting on the shore of a river. He is in the line of a 1.0 m long boat and is 5.5 m away from the centre of the boat. He wishes to throw an apple into the boat. If he can throw the apple only with a speed of 10 m/s, find the minimum and maximum angles of projection for successful shot. Assume that the point of projection and the edge of the boat are in the same horizontal level.

Sol:

Given: u = 10 m/s; max range, R_max = 6 m; min range, R_min = 5 m;

We have to find – θ_max and θ_min

We know, R = (u² sin 2θ)/g

R_max = (u² sin 2θ_max)/g → 6 = (10² sin 2θ_max)/10

Or, θ_max = ½ sin^-1 (0.6) 
Or, θ_max = 18.2⁰ or ½ (180- sin^-1 (0.6)) = 71.8⁰

R_min = (u² sin 2θ_min)/g → 5 = (10² sin 2θ_max)/10

Or, θ_max = ½ sin^-1 (0.5)
Or, θ_max = 15⁰ or ½ (180- sin^-1 (0.5)) = 75⁰ 
So, for a successful shot, θ may vary from 15° to 18° or 71° to 75°.

46. A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10 m/s with respect to the water, in a direction perpendicular to the river, (a) Find the time taken by the boat to reach the opposite bank, (b) How far from the point directly opposite to the starting point does the boat reach the opposite bank?

Sol:

V_{b, r} = velocity of boat w.r.t river.

V_{r, g} = velocity of river w.r.t ground.

V_{b, g} = velocity of boat w.r.t ground.

We have, V_{b, g} = V_{b, r} + V_{r, g}           ----- 1

(a) Taking vertical component of the equation 1

V_{b, g} sin θ = V_{b, r} + V_{r, g} cos 90⁰ = 10 m/s

The time taken by the boat to reach the opposite bank 

= (vertical displacement)/ (vertical velocity)

= 400/10 = 40 s.

(b) Taking horizontal component of the equation 1

V_{b, g} cos θ = V_{b, r} cos 90⁰ + V_{r, g} = 2 m/s

Horizontal displacement, AB

  = Horizontal component of resultant * time taken by the boat = 2 * 40 = 80 m.

47. A swimmer wishes to cross a 500 m wide river flowing at 5 km/h. His speed with respect to water is 3 km/h. (a) If he heads in a direction making an angle θ with the flow, find the time he takes to cross the river, (b) Find the shortest possible time to cross the river.

Sol:

(a)

V_{m, r} = velocity of boat w.r.t river.

V_{m, g} = velocity of river w.r.t ground.

V_{m, g} = velocity of boat w.r.t ground.

We have, V_{m, g} = V_{m, r} + V_{r, g} --------- (1)          

Taking vertical component of the equation 1

V_{m, g} sin 𝛂 = V_{m, r} sin θ + V_{r, g} cos 90⁰ = (3 sin θ) km/h

The time taken by the boat to reach the opposite bank

= (vertical displacement)/ (vertical velocity)

= 0.5/ (3 sin θ)hrs.= (10/sin θ) minute.                 

(b) For shortest possible time to cross the river, the velocity of man W.r.t. River should be perpendicular to the flow.

Therefore, the shortest possible time is = 0.5/3 hrs. = 10 minute.

48. Consider the situation of the previous problem. The man has to reach the other shore at the point directly opposite to his starting point. If he reaches the other shore somewhere else, he has to walk down to this point. Find the minimum distance that he has to walk.

Sol:

Velocity of man, V_m = 3 km/h

AB horizontal distance for resultant velocity R.

X-component of resultant 

R_x = 5 + 3 cos θ

t = 0.5 / 3sin θ

Which is same for horizontal component of velocity.

H = AB = (5 + 3 cos θ) (0.5 / 3 sin θ) 

              = (5 + 3 cos θ)/ (6 sin θ)

For H to be min (dH/d θ) = 0

→ d/dθ {(5 + 3 cos θ)/ (6 sin θ)} = 0

Or, –18 (sin² θ + cos² θ) – 30 cos θ = 0

Or, –30 cos θ = 18 → cos θ = –18 / 30 = –3/5

Sin θ = √ (1 - cos² θ) = 4/5

∴ H = (5 + 3 cos θ)/ (6 sin θ) = 2/3 km.

49. An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. A wind is blowing due north at a speed of 20 m/s. The air-speed of the plane is 150 m/s. (a) Find the direction in which the pilot should head the plane to reach the point B. (b) Find the time taken by the plane to go from A to B.

Sol:

Given: distance, AB = 500 km = 5 * 10⁵ m; speed of wind =20 m/s; speed of plane = 150 m/s;

In ∆ACD according to sine formula

We have, (CD)/sin A = (AC)/sin D

Or, sin A = (20 * sin 30⁰)/150 = 1/15

Or, A = sin^-1 (1/15) = 3.82⁰

(a) The direction of the plane to reach the point B should be sin^-1 (1/15) = 3.82⁰ east of the line AB.

(b)
θ = 30⁰ + 3.82⁰ = 33.82⁰

R = √ (150² + 20² + 2*150 * 20 cos 33.82⁰)

              = 167 m/s.

Therefore, the time taken by the plane to go from A to B is

= distance/speed 
= (5 * 10⁵)/167 = 2994 s = 50 min.

50. Two friends A and B are standing a distance x apart in an open field and wind is blowing from A to B. A beats a drum and B hears the sound t₁ time after he sees the event. A and B interchange their positions and the experiment is repeated. This time B hears the drum t₂ time after he sees the event. Calculate the velocity of sound in still air v and the velocity of wind u. Neglect the time light takes in travelling between the friends.
Sol:

In the first case, resultant velocity of sound = v + u

Therefore, (v + u) * t₁ = x

Or, (v + u) = x/t₁    ----------------- 1

In the 2nd case, resultant velocity of sound = v - u

Therefore, (v - u) * t₂ = x

Or, (v - u) = x/t₂    ----------------- 2

Solving equation 1 and 2 we get,

→ v = (x/2) [(1/ t₁) + (1/ t₂)] and

→ u = (x/2) [(1/ t₁) - (1/ t₂)]

51. Suppose A and B in the previous problem change their positions in such a way that the line joining them becomes perpendicular to the direction of wind while maintaining the separation x. What will be the time lag B finds between seeing and hearing the drum beating by A?

Sol:

Velocity of sound v, velocity of air u

Velocity of sound be in direction AC so it can reach B with resultant velocity AD.

Angle between v and u is θ > π/2.

Resultant AD = √ (v² - u²)

Here time taken by light to reach B is neglected. So time lag between seeing and hearing =time to hear the drum sound.

t = (velocity/Displacement) 
  = x / {√ (v² - u²)}

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