Question

A bus running at a speed of 18 km/h is stopped in 2.5 seconds by applying brakes. Calculate the retardation produced
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Answer 1

Speed of bus = Initial velocity = 18 kmph = 18*1000/60*60 = 30/6 = 5 m/s

Time = 2.5 s

Final velocity = 0

According to First equation of motion

A = V-U/T

A = 0-5/2.5

A = -5/2.5

A = -2 m/s2

Retardation = 2 m/s2

Answer 2

Answer:-

Given:-

• Initial velocity (u) = 18 km/h,

i.e. 5 m/s. [Multiply the speed value with 5/18]

• Final velocity (v) = 0 m/s [Stationary]
• Time taken (t) = 2.5 seconds

To Find: Deceleration produced.

We know,

a = (v - u)/t

where,

• a = Acceleration,
• v, u = Final and Initial velocities respectively,
• t = Time taken.

∴ a = - (5 m/s)/(2.5 s)

→ a = - 2 m/s²

∴ Retardation produced was (- 2 m/s²).