Question

A driver of a car moving at 20m/s finds a child on the road 50 m ahead and stops the car 10 m earlier
to the child. If the mass of the car with the driver is 1000 kg.
Calculate (a) the force exerted by the brakes on the car and
(b) the time taken to stop the car.
​

Answer 1

Force exerted by the brakes on the car  =5 kN  & Car stops in 4 sec   if Car moving at 20 m/s with 1000 kg weight stop 10 m before 50 m ahead child

Step-by-step explanation:

A driver of a car moving at 20m/s

child on the road 50 m ahead and car stop 10 m earlier to the child

=> Distance Covered by Car = 50 - 10 = 40 m

Initial Speed = 20 m/s

Final Speed = 0 m/s

using V² - U² = 2aS

=> 0² - 20² = 2 a * 40

=> a = - 5

Force =  ma

=> Force exerted by the brakes on the car  = 1000 * 5 = 5000N

= 5 kN

Time Taken to stop Car

V = U + at

=> 0 = 20 - 5 *t

=> t = 4 sec

Car stops in 4 sec

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Answer 2

Time Taken by Car to Stop is 4 seconds

Step-by-step explanation:

• Speed of the Car = 20 $ms^-^1$

• Distance between Road and Child = 50 m and car stopped 10 m before him

• Distance travelled by Car, S = 50 - 10 = 40 m
• The Initial Speed of Car, u = 20 $ms^-^1$
• The Final Speed of Car, V = 0 $ms^-^1$

using $V^2 - U^2 = 2aS$ (where 'V' is Final Speed, 'u' is Initial Speed, 'a' is Acceleration and 'S' is Distance)

$0^2 - 20^2 = 2 a \times 40$

 a = - 5

Force Applied as per Newtons law =  ma (Mass - Mass and a - Acceleration)

Force exerted on car through brakes  = $1000 \times 5$ = 5000N

= 5 kN

Time Taken to stop Car

V = U + at (I - Equation of Motion)

0 = 20 - $5 \times t$

 t = 4 sec

The car will be  stopped in 4 sec