Question

Let the position of an object moving along a straight line at any instant t is given by $s=t^3-3t^2-9t$, where s is in meters and t in seconds then

• A. The object is at the origin at three distinct instants t
• B. The velocity of the object is zero at t = 3s only
• C. The acceleration of the object is positive for all t > 2s
• D. For t < [0, 3], the maximum magnitude of the velocity is 12 m/s

Answer 1

The correct option is D For t < [0, 3], the maximum magnitude of the velocity is 12 m/s

$s=t^3-3t^2-9t$

$v=\frac{ds}{dt}$

$v=3t^2-6t-9$

$a=6(t-1)$

A: $s=0$

$t^3-3t^2-9t=0$

$t(t^2-3t-9)=0$

$t=0,t^2-3t-9=0$

$t=0$, $t=4.85,t=-1.85$

$t=-1.85$ not possible

so only 2 distinct t for which object at origin

B: $v=3t^2-6t-9$

$3t^2-6t-9=0$

$t^2-2t-3=0$

$t=-1,t=3$

$t=-1$ not possible

so $v=0$ only at $t=3$

C: $a=6(t-1)$

$a>0$

$6(t-1)>0$

$t>1$

D: $v=3t^2-6t-9$

roots $t=-1,t=3$

$\mid v\mid$ maximum at $t=\frac{-1=3}{2}=1s$

$\mid v{\mid }_{max}$=$\mid 3\times 1^2-6\times 1-9\mid$

$\mid v{\mid }_{max}$=$12$