Question

A particle moves along the x-axis. Its velocity-time graph is given below.
The distance travelled by the particle in 0 to $3s$

• A. Zero
• B. $4m$
• C. $8m$
• D. $12m$

Answer 1

The correct option is C $8m$

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)
$v=3t^2-6t$
From the graph we can see that the velocity changes its direction after t = 2 s
$\therefore \text{Distance}=\left|{[\frac{3t^3}{3}-\frac{6t^2}{2}]}_0^2\right|+\left|{[\frac{3t^3}{3}-\frac{6t^2}{2}]}_2^3\right|$
$|(8-0)-(12-0)|+|(27-8)-(27-12)|$
$=|-4|+|4|=8m$