Question

A car with speed 72km/hr suddenly applies break. The break has maximum ability to decelerate with
5ms-2. Find time taken to stop the car after applying breaks?

Answer 1

Answer:

The time taken (t) to stop the car is 4 seconds

Given:

1. Initial velocity (u) = 72 Km/hr.

2. Deceleration (a) = - 5 m/s²

3. Final velocity (v) = 0 m/s

Explanation:

____________________

First we need to convert Velocity into m/s (S.I Units)

we know,

⇒ 1 Km/h = 5 / 18 m/s

⇒ 72 Km/h = 72 × 5 / 18

                

⇒ 72 Km/h = 4 × 5

⇒ 72 Km/h = 20

⇒ v = 20 m/s

∴ We got the velocity in S.I units.

____________________

____________________

From First kinematic equation,

⇒ v = u + a t

Substituting the values,

⇒ 0 = 20 + (-5) × t

⇒ 0 = 20 - 5 × t

⇒ 0 = 20 - 5 t

⇒ 5 t = 20

⇒ t = 20 / 5

⇒ t = 4

⇒ t = 4 sec.

∴ The time taken (t) to stop the car is 4 seconds.

Note:

Symbols have their usual meanings

Hope this will help you....

Mark me as brain list...♂♂♂

Answer 2

Answer:  14.4 seconds

Explanation:

Initial velocity(u) = 72km/h

Final velocity (v) = 0km/h                         (since brakes are applied)

Acceleration (a) = 5m/s

WE KNOW,

a=v-u/t

⇒ t=v-u/a

⇒t= 0-72/-5

⇒t= -72/-5

⇒t= 14.4 s (ANS)