Question

Velocity of a particle as a function of time is shown in graph which is a parabola. If slope of the graph at $t=0$ is $2$ units and $v=p+qt+r{t}^2$, then (Assume SI units everywhere)

• A. $q=4$
• B. $r=-0.5$
• C. $v_0=6m/s$
• D. Velocity of particle is zero at $t=2\sqrt{3}+2$ sec

Answer 1

Answer: (B), (C), (D)

The correct options are
B $r=-0.5$
C $v_0=6m/s$
D Velocity of particle is zero at $t=2\sqrt{3}+2$ sec

Given $v=p+qt+r{t}^2$
at $t=0,v=4$ $\Rightarrow 4=p+0+0\Rightarrow p=4$ -- (1)
$a=\frac{dv}{dt}=q+2rt$
at $t=0,$ slope $=a=2$
$\Rightarrow 2=q+0$ and $q=2$ -- (2)
From graph, slope = 0 at t = 2
$\Rightarrow 0=2+2r\times 2\Rightarrow r=-\frac{1}{2}=-0.5$
$\therefore v=4+2t-\frac{1}{2}{t}^2$
Velocity at $t=2$
$v\left(2\right)=4+2\times 2+(-\frac{1}{2})\times 2^2=6m/s$
If velocity is zero,
$v=0=8+4t-t^2(t-2{)}^2=12t=2\sqrt{3}+2sec$