Question

Suppose ,if a body is moving with constant retardation and travels a certain distance ,where will its average velocity be equal to instantaneous velocity?

Answer 1

According to the question, the body is moving with a constant retardation. As such, its acceleration will be = -a.
If the body is initially moving with a velocity, u, due to the retardation its final velocity will be v = 0.
Let the time taken for the body to come to halt be t.
Now we know
$$\begin{gathered} v=u+at \\ \Rightarrow 0=u-at \\ \Rightarrow u=at.....\left(1\right) \end{gathered}$$

Also
$$\begin{gathered} y=ut-\frac{1}{2}a{t}^2 \\ \Rightarrow y=a{t}^2-\frac{1}{2}a{t}^2[\mathrm{Using}(1\left)\right] \\ \Rightarrow y=\frac{1}{2}a{t}^2.....\left(2\right) \end{gathered}$$

Here y is the distance travelled by the body in time t. If the body starts at a position, y₀ = 0, then average velocity of the body will be
$$\begin{gathered} \overline{v}=\frac{y-y_0}{t-0}=\frac{y}{t}=\frac{u^2}{2at}[\mathrm{from}(2\left)\right] \\ \overline{v}=\frac{u^2}{2u}=\frac{u}{2}[\mathrm{from}(1\left)\right].....\left(3\right) \end{gathered}$$


Instantaneous velocity of the body can be found as follows:
$$\begin{gathered} y=ut-\frac{1}{2}a{t}^2 \\ \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}t}=u-at.....\left(4\right) \end{gathered}$$

For average and instantaneous velocity to be equal, we equate (3) and (4)
$$\begin{gathered} \frac{u}{2}=u-at \\ \Rightarrow u=2at \\ \mathrm{Substituting}\mathrm{in}\left(2\right),\mathrm{we}\mathrm{find} \\ y=\frac{(2at{)}^2}{2a}=2a{t}^2 \end{gathered}$$

Thus required distance is
y = 2at²