Question

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

The particle changes its direction of acceleration, when

• A. $t<1s$
• B. $t=2s$
• C. $t>1s$
• D. $t\le 1s$

Answer 1

The correct option is C $t>1s$

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)
$v=3t^2-6t$
$\frac{dv}{dt}=6t-6$
For $t\le 1$, $\frac{dv}{dt}\le 0$
For $t>1$, $\frac{dv}{dt}>0$
Acceleration changing its direction after $1s$
$\therefore t>1s$
Acceleration at any point is the slope of $v-t$ graph at that point. From $t=0$ to $1s$, slope is negative but after $t=1s$, slope of $v-t$ graph becomes positive, which means direction of acceleration changes after $t=1s$.