Question

A car is travelling at 20m/s along a road A child runs out into the road 50m ahead and the car driver steps on brakes pedal. What must the car decceleration be if the car is stop just before it reaches the child

Answer 1

Answer:

4 m / sec² .

Explanation:

Given :

Initial velocity ( u ) = 20 m / sec

Distance ( s )  = 50 m

When car is stop just before reaches the child that means final velocity ( v )  = 0.

From third equation of motion we have

v² = u² + 2 a s

Putting values here we get

0² = 20² + 2 × 50 × a

0 = 400 + 2 × 50 × a

2 × 50 × a = - 400

a = - 4

Thus deacceleration must be 4 m / sec² to stop the car .

Answer 2

Que. A car is travelling at 20m/s along a road A child runs out into the road 50m ahead and the car driver steps on brakes pedal. What must the car decceleration be if the car is stop just before it reaches the child.

$\boxed{\boxed{\huge{\bf{Answer}}}}$

-4 m/s²

The information given in the question is that the speed of the car is 20m/s and the distance the car has to travel will be 50 m and after 50 m the velocity of the car must be 0 m/s.

initial velocity of the car (u) = 20 m/s

final velocity of the car (v) = 0 m/s

displacement (s) = 50 m

After analysing the quantities given to us we have to apply the third equation of motion

i.e. v² - u² = 2as

Now ,

(0)² - (20)² = 2 × a × 50

-400 = 100a

a = $\bf{\frac{-400}{100}}$

$\bf{\boxed{a = -4}}$