Question

What is retardation? Write its formula. Give an example where it acts on a body.
(A). A train starting from rest attained a velocity of $30m{s}^{-1}$ in 30 minutes. What is its acceleration?
(B). What is the distance travelled during this time?

Answer 1

Hint: When retardation acts on a body, it gradually slows down its speed so we can say that retardation is the opposite of acceleration. It is also a vector quantity. The acceleration of the train can be calculated by substituting the corresponding values in the formula for acceleration, while the distance travelled can be calculated by substituting corresponding values into the formula for speed.
Formulas used:
$r=-{\displaystyle \frac{(v-u)}{t}}$
$s={\displaystyle \frac{d}{t}}$

Complete answer:
The change in velocity per unit time is called acceleration. Its SI unit is $m{s}^{-2}$. It is a vector quantity.
When the acceleration applied is opposite to the motion of the body, it is called retardation. The formula for retardation is-
$\begin{array}{rl}& r=-a\\ & \Rightarrow r=-{\displaystyle \frac{(v-u)}{t}}\end{array}$
Here, $r$ is the retardation
$a$ is the acceleration
$v$ is the final velocity
$u$ is the initial velocity
$t$ is the time taken
Magnitude of retardation is negative of acceleration as it acts in the opposite direction of acceleration.
(A). Given, for a train, $u=0$, $v=30m{s}^{-1}$, time taken-
$\begin{array}{rl}& t=30min\\ & \Rightarrow t=30\times 60\sec \\ & \therefore t=1800\sec \end{array}$
We know,
$a={\displaystyle \frac{v-u}{t}}$
In the above equation, we substitute given values to get,
$\begin{array}{rl}& \Rightarrow a={\displaystyle \frac{30-0}{1800}}m{s}^{-2}\\ & \Rightarrow a=0.02m{s}^{-2}\end{array}$
Therefore, the acceleration of the train is $0.02m{s}^{-2}$.
(B). We know that speed is the distance travelled in unit time. It is given by-
$s={\displaystyle \frac{d}{t}}$
Here, $s$ is speed
$d$ is the distance travelled
$t$ is the time taken
Given, $s=30m{s}^{-1}$, $t=1800s$
In the above equation, we substitute the given values to get,
$\begin{array}{rl}& 30={\displaystyle \frac{d}{1800}}\\ & \Rightarrow d=30\times 1800\\ & \Rightarrow d=54\times {10}^{3}m=54\times {10}^3\times {10}^{-3}km\\ & \therefore d=54km\end{array}$
Therefore, the distance travelled by the train in the time is $54km$.

Therefore, the train accelerates at $0.02m{s}^{-2}$ and travels $54km$.

Note:
The equations of motion in one direction can only be applied when acceleration is constant. Acceleration comes into play when an external force is acting on a body. Speed is a scalar quantity. It is the magnitude of velocity. While distance is a scalar quantity, It is the magnitude of displacement.