Question

A car running at 72 km h$^{-1}$ is slowed down to 18 h$^{-1}$ by the application of braokes over s distance of 20 m. Calculate de - acceleration of car. If the brokes are applied with the same force calculate (i) total time in which car stops and (ii) total distance covered by it.​

Answer 1

Answer:

A car running at a speed of 72 km/hr is slowed down to 18 km/hr over a distance of 40 m. Calculate :(1) the retardation produced by the brakes.

Answer 2

Given:-

u = 72 km h$^{–1}$ = 20 ms$^{–1}$;

v = 18 km h$^{–1}$ = 5 ms$^{–1}$

S = 20 m

where u means initial velocity;

v means final velocity and

S means distance

Case (i)

Using third equation of motion

v$^{2}$ – u$^{2}$ = 2as

➜ (5)$^{2}$ – (20)$^{2}$ = 2 × a × s

➜ a = –375/40

= –9.375 ms$^{–2}$

We know that,

De - acceleration = (–a) = –(–9.375 ms$^{–2}$) = 9.375 ms$^{–2}$.

In Case (ii)

u (initial velocity) = 72 km h$^{–1}$;

v (final velocity) = 0;

a (acceleration) = –9.375 ms$^{–2}$

Total time in which car stops:

Using first equation of motion

v = u + at

➜ 0 = 20 – 9.375 × t

➜ t = 2.13s.

Total distance covered:

Using second equation of motion

S = ut + 1/2 at$^{2}$

= 20 × 2.13 –1/2 × 9.375 × (2.13)$^{2}$

= 21.34 m.

Thus, total time in which the car stops = 2.13s

and the total distance covered by it = 21.34 m.