Question

A car is moving at a constant velocity of 36 km/h.On applying brakes it stops at a distance of 10m.Calculate the deceleration and time taken by car to stop.​

Answer 1

Answer:

Initial velocity(u) = 36 km/h

= 10 m/s

final velocity(v) = 0

distance covered after the

break applied(s) = 10 m

now,

v² = u² + 2as

=> a= (v² - u²)/2s

=> a= (0² - 10²)/2 X 10

=> a= -100/20

=> a= -5 m/s²

hence, the car decelerate at the rate of 5 m/s² , here the negative us denoting that the car is decelerating.

now,

v= u + at

=> t= (v-u)/a

=> t= (0-10)/(-5)

=> t= 10/5

=> t= 2 s

hence, it will take 2 s to stop.

Answer 2

Given :

• Initial velocity (u) = 36 km/hr
• Final velocity (v) = 0 km/hr

(As, brakes are applied)

• Distance (s) = 10 m

Find :

Deceleration (-a) and time (t) taken by car.

Solution :

We have that..

• v² - u² = 2as

(Third equation of Motion)

Before putting the values, first convert the initial and final velocities into m/s.

To convert km/hr into m/s. Multiply with 5/18.

So,

Initial velocity = 36 × 5/18

=> 10 m/s

Final velocity = 0 × 5/18

=> 0 m/s

Now, put the values in above formula

=> (0)² - (10)² = 2a(10)

=> - 100 = 20a

=> a = -1/2

=> a = - 5 m/s²

∴ Deceleration is 5 m/s².

Now,

• v = u + at

(First equation of motion)

Put the known values in above formula

=> 0 = - 10 + (5)t

=> 10 = 5t

=> t = 10/5

=> t = 2 sec

∴ Time taken by car to stop is 2 sec.