A Bus Running at a Speed of 18 km/hr is stopped in 2.5 sec by applying brakes. Calculate the Retardation Produced

By BYJU'S Exam Prep

Updated on: September 25th, 2023

The required retardation is 2 m/s²

Acceleration is the rate at which velocity changes in relation to time. Since acceleration has both a magnitude and a direction, it is a vector quantity. Additionally, it is the first derivative of velocity with respect to time or the second derivative of position with respect to time.

Retardation is the rate at which velocity decreases over time.

Consider the case where a car is slowing down and its acceleration is negative because of the decline in velocity.

But retardation has certain benefits.

Acceleration is a negative of retardation.

Step 1: List the given values

Given, the initial velocity of the bus, v_i = 18 km/hr

We know that 1km/hr = 5/18 m/s

18 km/hr = (5/18 x 18) m/s

= 5 m/s

Since the bus comes to a stop, its final velocity is equal to zero.

The time taken t = 2.5 s

Table of content

1. A Bus Running at a Speed of 18 km/hr is stopped in 2.5 sec by applying brakes. Calculate the Retardation Produced

Step 2: Now we have to calculate the retardation

We know that acceleration,

a = (vf – vi)/t

Substituting the values we get

= (0 – 5)/ 2.5

= -2 m/s²

We are aware that the opposite of acceleration is retardation.

a_r = -a

a_r = 2 m/s²

Hence, the required retardation is 2 m/s²

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