Question

A particle moves along the x-axis. Its velocity-time graph is given below.

The displacement of the particle in $0$ to $3s$

• A. Zero
• B. 4 m
• C. 8 m
• D. 12 m

Answer 1

The correct option is A Zero

As the curve is parabolic let us assume a second order equation for velocity to fit in this curve
Let $\begin{array}{}v=a{t}^2+bt+c& \text{(1)}\end{array}$
Apply boundary condition to above equation to find a,b and c
At $t=0s,v=0m{s}^2$
$0=(a\times 0^2)+(b\times 0)+c$
$c=0$
At $t=1s,v=-3m{s}^2$
$-3=(a\times 1^2)+(b\times 1)+0$
$\begin{array}{}-3=a+b& \text{(2)}\end{array}$
At $t=2s,v=0m{s}^2$
$0=(a\times 2^2)+(b\times 2)$
$\begin{array}{}b=-2a& \text{(3)}\end{array}$
Substitute (3) in equation (2) we get,
$-3=a-2a$
$a=3$
Substitute value of $a$ in equation (2)
$-3=3+b$
$b=-6$
Substitute these $a,b$ and $c$ in equation (1)
$v=3t^2-6t$

Displacement is given as the area under the curve i.e.
Displacement $={\int }_0^{3}vdt$
$={\int }_0^3(3t^2-6t)dt$
$={[\frac{3t^3}{3}-\frac{6t^2}{2}]}_0^3\Rightarrow (27-0)-3(9-0)$
Displacement = 0