Ques 2 and 3

Dear Student,

Please find below the solution to the asked query:

Question (2)

Given equation for the position of the particle is,

$x=u\left(t-2\right)+a{\left(t-2\right)}^2$
So, the velocity and acceleration equations are,

$$\begin{gathered} x=u\left(t-2\right)+a{\left(t-2\right)}^2 \\ v=\frac{dx}{dt}=\frac{d}{dt}\left\{u\left(t-2\right)+a{\left(t-2\right)}^2\right\}=u+2a\left(t-2\right) \\ a=\frac{dv}{dt}=\frac{d}{dt}\left\{u+2a\left(t-2\right)\right\}=2a \\ So, at t=0 \\ v=u-4a\ne u \end{gathered}$$

at t = 2 s,

$$\begin{gathered} x=u\left(t-2\right)+a{\left(t-2\right)}^2\Rightarrow x=u\left(2-2\right)+a{\left(2-2\right)}^2 \\ \Rightarrow x=0 \end{gathered}$$

Therefore Options (c & d) are correct.

Question (3)

Here Options (a,b & c) are correct.

As we know the displacement is the shortest possible distance between two points. In the same way the magnitude of the average velocity is the minimum possible value for the average speed but the average speed cannot be less than that of the magnitude of average speed.

If the particle moves in a circle its speed may constant but its velocity is not going to be constant.

If the particle comes to its initial position than the average velocity becomes zero, it is not mandatory that the instantaneous velocity should become zero (consider the first statement).
 

Hope this information will clear your doubts about the topic.

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