Question

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

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

Answer 1

Answer: (A)

Given,
Displacement of the body at time '$t$'
$x=4t+5t^3\dots (i)$
$\therefore t=1s$
$x_1=4\times 1+5(1{)}^3=9$
and $t=2s$,
$x=4\times 2+5\times (2{)}^3$
$=48$
$\therefore$ Displacement of the body from $t=1s$ to $t=2s$,
$\Delta x=x_2-x_1=48-9=39$
$\therefore v_{avg}=\frac{\Delta x}{\Delta t}=\frac{39}{2-1}$
$=39m/s$
From Eq. (i),
$x=4t+5t^3$
$\therefore \frac{dx}{dt}=V=4+15t^2$
$v_{inst}$ at $t=1s$
$\therefore V_{inst}=4+15\times (1{)}^2$
$=19m/s$
Thus, $V_{avg}-V_{inst}=39-19$
$=20m/s$