Question

A stone thrown vertically upwards rises ‘$s$’ metre in $t$ seconds, where $s=80t-16t^2$, then the velocity after $2$ second is

• A. $8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$
• B. $16\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$
• C. $32\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$
• D. $64\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$

Answer 1

The correct option is B

$16\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$

Step1: Given data.

Equation of motion of the stone moving vertically upward, $s=80t-16t^2$

Step2: Finding the the velocity after $2$ second

We have,

$s=80t-16t^2$ .…..$\left(i\right)$

On differentiating equation $\left(i\right)$ both sides we get,

$\Rightarrow$$\frac{ds}{dt}=\frac{d\left(80t-16t^2\right)}{dt}$

$\Rightarrow$$\frac{ds}{dt}=80-32t$

Where $\frac{ds}{dt}$ is the velocity of the stone.

Now, at $t=2s$ the velocity will be,

$\frac{ds}{dt}=80-32\times 2$

$\Rightarrow$$\frac{ds}{dt}=16$

$\because$the velocity after $2$ second is $16\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$.

Hence, option B is correct.