Question

A particle moves so that $S=6+48t-t^3$. The direction of motion reverses after moving a distance of

• A. $63$
• B. $104$
• C. $134$
• D. $288$

Answer 1

The correct option is C

$134$

Step 1: Given data.

Distance covered by particle,$S=6+48t-t^3$

Step 2: Finding the distance after the direction of motion reverses.

Since, we have,

$S=6+48t-t^3$

We know that,

$v=\frac{dS}{dt}$ $\left[\because v=velocity\right]$

$v=48-3t^2$ …….$\left(i\right)$

Now, when the direction reversed then.

$v=0$

$48-3t^2=0$ $\left[\because Fromequation\left(i\right)\right]$

$t=-4$ or $t=4$

Again,

At $t=4$.

$S=6+48\times 4-{\left(4\right)}^3$

$S=134$

Hence, option C is correct.