Question

A car is moving with a speed of 54 km / h. If after 3 s, the driver applies brakes and it stops, then how much distance is covered by the car before coming to rest?

• A. 22.5 m
• B. 20 m
• C. 25 m
• D. 45.2 m

Hint:

When the vehicle is stopping, use the equations of motion for uniformly accelerated motion. Remember to take acceleration as negative when decelerating.

Answer 1

The Correct Option is A

First, convert the speed to m/s: \[ \text{Speed} = \frac{54 \times 1000}{3600} = 15 \text{ m/s} \] The car stops after 3 seconds, so the distance covered can be calculated using the equation of motion: \[ v = u + at \] Since the car stops, final speed \( v = 0 \), initial speed \( u = 15 \text{ m/s} \), and time \( t = 3 \text{ s} \). From the equation: \[ 0 = 15 + a(3) \] Solving for \( a \): \[ a = \frac{-15}{3} = -5 \text{ m/s}^2 \] Now, using the equation for distance covered under uniform acceleration: \[ s = ut + \frac{1}{2}at^2 \] Substitute the values: \[ s = 15 \times 3 + \frac{1}{2}(-5)(3^2) \] \[ s = 45 - 22.5 = 22.5 \text{ m} \] Thus, the distance covered before the car comes to rest is 22.5 m.