Question

A dynamite blast blows a heavy rock straight up with a launch velocity of 160 m/sec. It reaches a height of $s=160t-16t^2$ after t sec. The velocity of the rock when it is 256 m above the ground on the way up is
(a) 98 m/s
(b) 96 m/s
(c) 104 m/s
(d) 48 m/s

Answer 1

Hint: Firstly, we have to find the velocity by differentiating $s=160t-16t^2$ with respect to t. Then, find the value of t when $s=256m$ by substituting it in the equation $s=160t-16t^2$ . Substitute these values of t in the velocity equation and simplify.

Complete step by step answer:
We are given that the velocity with which the rock is launched is 160 m/sec. We are given that the rock reaches the height $s=160t-16t^2$ . This means that $s=160t-16t^2$ is the distance.
We know that velocity is the derivative of distance with respect to time.
$\Rightarrow v={\displaystyle \frac{ds}{dt}}$
We know that ${\displaystyle \frac{d}{dx}}{x}^n=n{x}^{n-1}$ .
$\Rightarrow v\left(t\right)=160-32t...\left(i\right)$
We have to find the velocity when $s=256\text{ m}$ . Let us find t from $s=160t-16t^2$ by substituting this value of s.
$\begin{array}{rl}& \Rightarrow 256=160t-16t^2\\ & \Rightarrow 16t^2-160t+256=0\\ & \Rightarrow 16(t^2-10t+16)=0\\ & \Rightarrow t^2-10t+16=0\end{array}$
Let us factorize the above polynomial by splitting the middle term.
$\begin{array}{rl}& \Rightarrow t^2-2t-8t+16=0\\ & \Rightarrow t(t-2)-8(t-2)=0\\ & \Rightarrow (t-8)(t-2)=0\\ & \Rightarrow (t-8)=0,(t-2)=0\\ & \Rightarrow t=8\text{ sec },2\text{ sec}\end{array}$
Now, let us find the value of $v\left(t\right)$ at $t=8\text{ and }t=2$ .
$\begin{array}{rl}& \Rightarrow v\left(8\right)=160-32\times 8\\ & \Rightarrow v\left(8\right)=160-256\\ & \Rightarrow v\left(8\right)=-96\text{ m/s}\end{array}$
Now, we have to find $v\left(t\right)$ at $t=2$ .
$\begin{array}{rl}& \Rightarrow v\left(2\right)=160-32\times 2\\ & \Rightarrow v\left(2\right)=160-64\\ & \Rightarrow v\left(2\right)=96\text{ m/s}\end{array}$
Therefore, the velocity of the rock when it is 256 m above the ground on the way up is 96 m/s.

So, the correct answer is “Option b”.

Note: Students must note that derivative of distance is velocity and derivative of velocity is acceleration. Acceleration is also the second order derivative of distance.
$\Rightarrow a={\displaystyle \frac{dv}{dt}}={\displaystyle \frac{d^{2}s}{d{t}^2}}$
Students must be thorough with the formulas of derivatives.