Question

A scooter moving at a speed of 16 m/s is stopped by applying brakes which produces an uniform retardation of 0.5 metre per second square . How much distance will be covered by the scooter before it stops?

Answer 1

Answer:

The speed of the scooter = 16 m/s.

The retardation of the body = -0.5 m/s².

We need to find the Distance travelled by the scooter before it stops.

Using third equation of motion :

⇒ v² - u² = 2as

⇒ 0² - 16² = 2×(-0.5)×s

⇒ -256 = (-1)×s

⇒ -256 = -s

⇒ s = 256 m

Hence, the Distance travelled by the scooter before it stops = 256 m.

More information:

• Retardation is the negative acceleration of the body.
• Three equations of motion:

1. v = u + at
2. s = ut + ½at²
3. v² - u² = 2as

• Speed is a scalar quantity, whereas velocity is a vectore quantity.
• Speed is the rate of change of distance.
• Velocity is the rate of change of displacement.

Answer 2

Given :-

• Speed Of Scooter = 16m/s.

• Negative Acceleration = -0.5m/s²

To Find :-

• Distance covered by the scooter before it stops.

Solution :-

We know the 3rd Equation Of Kinematics,

v² - u² = 2as.

Substitute The Values.

➛0² - 16² = 2(-0.5)s.

➛-256 = -s.

➛s = 256

Hence, The Distance Covered by the Scooter before it stops is 256 metres.