Question

A car running at 5400 km/h is slowed down to 1800 km/h in 21 seconds by the
application of brakes.Calculate the retardation produced.

Answer 1

Retardation - Motion

Acceleration/Retardation is defined as the rate of change in velocity with respect to time. It is measured as metre per second square. In Mathematically,

$\boxed{\bf{\bar{a} = \dfrac{\Delta v}{\Delta t}}}$

Where, ā denotes acceleration, Δv denotes change in velocity and Δt denotes time taken or time intervals.

As per the question, A car is running at 5400 km/h is slowed down to 1800 km/h in 21 seconds by applying the brakes. So from here we can conclude that the initial velocity is 5400 km/h and the final velocity of the train is 1800 km/h.

In order to find the retardation, first we have to convert the units of initial and final velocity from km per hour to m per second.

We know that, to convert the unit from km per hour to m per second we have to divide the velocity value by 3.6. [You can also multiply the velocity value by 5/18 to convert the velocity from km per hour to m per second]

Initial velocity, u = 5400/3.6 = 1500m/s

Final velocity, v = 1800/3.6 = 500m/s

Now we know that, Retardation is the rate of change in velocity with respect to time. Therefore,

$\implies \bar{a} = \dfrac{\Delta v}{\Delta t} \\ \\ \implies \bar{a} = \dfrac{v-u}{t}$

Now by substituting the known values in the above equation/formula, we get:

$\implies \bar{a} = \dfrac{500 - 1500}{21} \\ \\ \implies \bar{a} = \dfrac{-1000}{21} \\ \\ \implies \boxed{\bf{\bar{a} = -47.61}}$

The negative acceleration is also referred to as retardation and the train is said to be retarding.

Hence, the retardation produced by the train is -47.61m/s².

Answer 2

Retardation

Retardation refers to act of delay. When velocity decreases, it's acceleration is negative. The Negative acceleration is known as Retardation. It's SI unit is m/s².

According to the question :

Firstly Let's find the Initial Velocity $u$ and Final Velocity $v$. Converting km/hr into m/s by dividing the value by 3.6,

• Initial Velocity $u$ = $\sf{\dfrac{5400}{3.6}}$ = $\sf{1500}$
• Final Velocity $v$ = $\sf{\dfrac{1800}{3.6}}$ = $\sf{500}$
• Time = $\sf{21\:seconds}$

We know that,

$\boxed{Final\:Velocity=Initial\: Velocity\:-\:Acceleration\:\times\:Time}$

where,

• Final Velocity = ${v}$,
• Initial Velocity = ${u}$,
• Acceleration = ${a}$ and
• Time = ${t}$.

Substituting values, we get :

$\sf\implies{v\:=\:u\:-\:at}$

$\sf\implies{1500\:=\:500\:-\:a\:\times\:21}$

$\sf\implies{1500\:=\:500\:-\:21a}$

$\sf\implies{21a\:=\:500\:-\:1500}$

$\sf\implies{21a\:=\:-1000}$

$\sf\implies{21a\:=\:-1000}$

$\sf\implies{a\:=\:{\dfrac{-1000}{21}}}$

$\sf\implies{a\:=\:{\cancel{\dfrac{-1000}{21}}}}$

$\sf\implies{a\:=\:-47.61}$

Hence, the retardation produced = -47.61 m/s².