Question

Question 3. 6. A car moving along a straight highway with speed of 126 km h-1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

Answer 1

Answer :

the time taken by the car = 11.4s

Answer 2

Given :-

Distance covered by the car before coming to rest = 200 m

A car moving along a straight highway with speed = 126 km h⁻¹

To Find :-

Time taken for the car to stop.

Analysis :-

First, using the third equation of motion, find the acceleration.

Using the first equation of motion substituting the value given in the question accordingly considering the time as a variable.

Solution :-

We know that,

• v = Final velocity
• a = Acceleration
• u = Initial velocity
• t = Time
• s = Distance

By the third equation of motion,

$\underline{\boxed{\sf Third \ equation \ of \ motion=v^2 - u^2 = 2as }}$

Given that,

Initial velocity (u) = 126 km/h = 35 m/s

Distance travelled (s) = 200 m

Final velocity (v) = 0 (Since the car comes under rest)

Substituting their values,

a = (v² - u²)/2s

a = (0 - 35)/2×200

a = -1225/400

a = -3.06 m/s²

Therefore, the acceleration is -3.06 m/s².

Let the time be 't'.

Using first equation of motion,

$\underline{\boxed{\sf First \ equation \ of \ motion= v=u+at}}$

Given that,

Initial velocity (u) = 126 km/h = 35 m/s

Distance travelled (s) = 200 m

Final velocity (v) = 0

Acceleration = -3.06 m/s²

Substituting their values,

0 = 35 - 3.06 × t

t = 35/3.06

t = 11.43 sec

Therefore, it takes 11.43 sec for the car to stop.