Question

A particle starts its motion from point A in the positive x direction in such a way that after every t sec it is at a distance of $t^2$ from the starting point. Find its instantaneous velocity at $t=4^{th}$ sec also the rate at which velocity changes.

• A. v = 8 m/s rate of velocity change = 2
• B. v = 16 m/s rate of velocity change = 4
• C. v = 8 m/s rate of velocity change = 8
• D. v = 16 m/s rate of velocity change = 16

Answer 1

The correct option is A

v = 8 m/s rate of velocity change = 2

Given at any time the distance of the particle from origin will be equal to $t^2$

In other words x = $t^2$

To find instantaneous velocity we need $\frac{dx}{dt}$

$\Rightarrow x\left(t\right)=t^2$

$\frac{dx}{dt}=2t$

v = 2t

So at t = 4 sec

v = 2 \times 4 = 8 m/s

Rate at which velocity changes

$\frac{V_2-V_1}{t_2-t_1}$

Let's take $t_1=0$

$t_2=4$

V at $t_2=8m/s$

So rate = $\frac{8-0}{4-0}=2$

Also rate can be found by $\frac{dv}{dt}$

V = 2t

So $\frac{dv}{dt}=2$