Solutions to Exercises on Kinematics - Rest And Motion HC Verma's Concepts of Physics Part 1 (17-21)

 Q#17

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

Answer:
First make the units uniform, distance x = 5.0 cm = 0.05 m. Now from the equation of accelerated motion v² = u² + 2ax 

0² = 350² + 2(a)0.05

a = -350²/0.1 = -1225000 m/s² (-ve sign is for retardation)

So, retardation = 12.2 x 10⁵ m/s²


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

Answer:
Given u = 0, t = 5.0 s, v =18 km/h = 18000/3600 m/s = 5 m/s

(a) From v = u + at

5 = 0 + a(5)

a = 1 m/s²

From the equation x = ut + ½at²

Average velocity x/t = u + ½at = 0+ ½(1)5 = 2.5 m/s

(b) The distance traveled by the particle during this period = Average_velocity x time taken

= 2.5 m/s (5 s) = 12.5 m

Q#19
A driver takes 0.2 s to apply the brakes after he sees a need for it. This is called the reaction time of the 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 traveled by car after he sees the need to put the brakes on.    

Answer:
u = 54 km/h = 54000/3600 = 15 m/s

First the distance traveled during reaction time = speed x time = 15 m/s x 0.2 s = 3 m

Now for the decelerated motion u = 15 m/s, v = 0 m/s, a = -6.0 m/s²

From v² = u² + 2ax

0² = 15² + 2(-6)x 

x = 225/12 m = 18.75 m

So total distance traveled by the car after the driver sees the need to put the brakes on = 18.75 + 3 = 21.75 m ≈ 22 m

Q#20
Complete the following table:

Answer:
Braking distance = (u² - v²)/2a = u²/2a 
Distance in reaction time = ut
Total stopping distance TSD = ut + u²/2a

Now for u = 54 km/h = 15 m/s, t = 0.2 s Braking distance a = 18.75 m ≈19 m
TSD b = 22 m

But for t = 0.3 s and u = 72 km/h = 20m/s , Braking distance c = 20²/(2 x 6) = 400/12 = 33.33 m ≈ 33 m and TSD d = 20(0.3) + 33 = 39 m

Now for deceleration 7.5 m/s², t = 0.2 s, u = 15m/s, Braking distance e = 15²/(2 x 7.5) = 225/15 = 15 m and TSD f = 15(0.2) + 15 = 18 m

Again for deceleration 7.5 m/s², t = 0.3 s, u = 20m/s, Braking distance g = 20²/(2 x 7.5) = 400/15 = 26.67 m ≈ 27 m, and TSD h = 20 x 0.3 + 27 = 33 m.

Now it can be filled in given table as below:

Car Model	Driver X Reaction time 0.20 s	Driver Y Reaction time 0.30 s
A (deceleration on hard braking = 6.0 m/s2)	Speed = 54 km/h Braking distance a = 19 mTotal stopping distance b = 22 m Speed = 72 km/h Braking distance	Speed = 72 km/h Braking distance c = 33 m Total stopping distance d = 39 m
B (deceleration on hard braking = 7.5 m/s2)	Speed = 54 km/h Braking distance e = 15 m Total stopping distance f = 18 m	Speed = 72km/h Braking distance g = 27 m Total stopping distance h = 33 m

Q#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? 

Answer:
72 km/h = 72000/3600 m/s = 20 m/s

90 km/h = 90000/3600 m/s = 25 m/s

Suppose the jeep catches the bike at a distance of x m from the turning. Time taken by the bike in covering x m = x/20 s.

Since the jeep starts 10 s later, so it takes 10 s less time to cover the equal distance. So time taken by the jeep in covering x m = (x/20)-10 s. But from the jeep's speed, it comes to be x/25 s. These two times must be equal. Equating both times we get,

(x/20) – 10 = x/25,
(x – 200)/20 = x/25
(x – 200)/4 = x/5
5x – 1000 = 4x
x = 1.0 km

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