Question

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

126 km/h×5/18

=35 m/s

Using Equation v²=u²+2as

35²= 0+2×a×200

1225= 400a

a= 1225/400

a=3.0625 m/s²

a=3.06m/s²* Round off

Time= v=u+at

35=0+3.06×t

35/3.06=t

t=11.44 s

Answer 2

Given :

• Initial velocity (u) = 126 km/h
• Final Velocity (v) = 0 m/s
• Distance (s) = 200 m

To Find :

• Retardation of the car = ?
• Time taken by the car = ?

Solution :

First of all convert the unit of Initial velocity from km/h to m/s :

→ Initial velocity (u) = 126 km/h

→ Initial velocity (u) = 126 × 5/18 m/s

→ Initial velocity (u) = 7 × 5

→ Initial velocity (u) = 35 m/s

• Hence,the initial velocity of the car is 35 m/s.

Now, let's calculate the retardation of the car by using third equation of motion :

→ v² - u² = 2as

→ (0)² - (35)² = 2a × 200

→ 0 - 1225 = 400a

→ a = (-1225) ÷ 400

→ a = -3.06 m/s²

• Hence,the retardation of the car is 3.06 m/s².

Finding the time taken by the car to stop by using first equation of motion :

→ v = u + at

→ 0 = 35 + (-3.06) × t

→ 0 = 35 - 3.06t

→ 3.06t = 35

→ t = 35 ÷ 3.06

→ t = 11.437 seconds

• Hence,the time taken by the car to stop is 11.437 seconds.