Question

A stone thrown vertically upwards rises $'s'$ metres in $t$ seconds , where $s = 80t - 16{t^2}$ , then the velocity after $2$ second is
A) $8m.{\sec ^{ - 1}}$
B) $16m.{\sec ^{ - 1}}$
C) $32m.{\sec ^{ - 1}}$
D) $64m.{\sec ^{ - 1}}$

Answer 1

Hint: Velocity is the derivative of displacement with respect to time , also relation between distance and time where velocity is $\dfrac{d}{{time}}$ , where $d$ is the distance . First we find the derivative of $s$ with respect to time . After that use the given data $t = 2$ second and get the required answer . Differentiation of ${x^n}$ is $\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$ .

Complete step by step answer:
From the given data $s = 80t - 16{t^2}$
Now differentiating both sides of above equation and we get
$\dfrac{{ds}}{{dt}} = \dfrac{d}{{dt}}\left( {80t - 16{t^2}} \right)$
$ = 80 - 32t$
From the given data we do not know about the velocity we find this , so let $v$ be the velocity .
Therefore $v = \dfrac{{ds}}{{dt}}$
Put the value of $\dfrac{{ds}}{{dt}} = 80 - 32t$ in the above equation and we get
$ \Rightarrow v = 80 - 32t$
Now put the value of $t = 2$ in above equation , we get
$ \Rightarrow v = 80 - 32 \times 2$
$ \Rightarrow v = 80 - 64$
$ \Rightarrow v = 16$
Therefore the required velocity is $16m.{\sec ^{ - 1}}$. So, Option (B) is correct.

Note:
Differentiating is a process of finding the derivative or rate of changes of a function . We know the unit of velocity that is $m.{\sec ^{ - 1}}$ . If we forget this, we establish time by using the formula of velocity. The formula of unit velocity is $\dfrac{{unit\;of\;distance}}{{unit\;of\;time}}$. We know the unit of distance is metre and unit of time is second . Therefore the unit of velocity is $m.{\sec ^{ - 1}}$.