Question

A particle moves in a straight line so that it covered a distance $a{t}^3+bt+5$ metres in $t$ seconds. If its acceleration after $4$ seconds is $48\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$, then $a$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B

$2$

Step 1: Give data and assumptions.

Distance covered, $s=a{t}^3+bt+5$

Acceleration after 4 second, $a=48\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$

Step 2: Finding the value $a$ from the given equation.

Since,

$s=a{t}^3+bt+5$.

Differentiate the above equation with respect to time $t$ we get.

$\frac{ds}{dt}=\frac{d\left(a{t}^3+bt+5\right)}{dt}$

$\Rightarrow$$\frac{ds}{dt}=3a{t}^2+b$

$\Rightarrow$$v=3a{t}^2+b$ .$\left[\because \frac{ds}{dt}=v\right]$ .$\left(i\right)$

Now,

We know that,

$a=\frac{dv}{dt}$

Where $a$ is acceleration and $\frac{dv}{dt}$ is the change in velocity with respect to time.

$\Rightarrow$$48=\frac{d\left(3a{t}^2+b\right)}{dt}$ $\left[\because Fromequation\left(i\right)\right]$

$\Rightarrow$$a=\frac{48}{6t}$

At, $t=4$

$\Rightarrow$$a=\frac{48}{6\times 4}$

$\Rightarrow$$a=2$

Hence, option B is correct.