Question

The displacement of a body is given by $ x = 4 t + 5t^{3} $ , where $ x $ is in metre and $ t $ is in second. The difference between the average velocity of the body in the time-interval $ t = 1\, s $ to $ t = 2\, s $ and its instantaneous velocity at $ t = 1 \,s $ is

• A. $ 20.0 \, m/s $
• B. $ 22.5 \, m/s $
• C. $ 27.0 \, m/s $
• D. $ 39.0 \, m/s $

Answer 1

Answer: (A)

Given,
Displacement of the body at time '$t$'
$x=4t+5t^{3} \ldots(i)$
$\therefore t=1\,s$
$x_{1}=4\times 1+5(1)^{3}=9$
and $t=2s$,
$x=4\times 2+5\times (2)^{3}$
$=48$
$\therefore $ Displacement of the body from $t = 1s$ to $t=2s$,
$\Delta x=x_{2}-x_{1}=48-9=39$
$\therefore v_{avg} =\frac{\Delta x}{\Delta t}=\frac{39}{2-1}$
$=39 \,m/s$
From Eq. (i),
$x=4t+5t^{3}$
$\therefore \frac{dx}{dt}=V=4+15t^{2}$
$v_{inst}$ at $t=1\,s$
$\therefore V_{inst}=4+15\times(1)^{2}$
$=19\, m/s$
Thus, $V_{avg} - V_{inst}=39-19$
$=20\, m/s$