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Question: A dynamite blast blows a heavy rock straight up with a launch velocity of 160 m/sec. It reaches a he...

A dynamite blast blows a heavy rock straight up with a launch velocity of 160 m/sec. It reaches a height of s=160t−16t2s=160t-16{{t}^{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

Explanation

Solution

Firstly, we have to find the velocity by differentiating s=160t−16t2s=160t-16{{t}^{2}} with respect to t. Then, find the value of t when s=256ms=256m by substituting it in the equation s=160t−16t2s=160t-16{{t}^{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−16t2s=160t-16{{t}^{2}} . This means that s=160t−16t2s=160t-16{{t}^{2}} is the distance.
We know that velocity is the derivative of distance with respect to time.
⇒v=dsdt\Rightarrow v=\dfrac{ds}{dt}
We know that ddxxn=nxn−1\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} .
⇒v(t)=160−32t...(i)\Rightarrow v\left( t \right)=160-32t...\left( i \right)
We have to find the velocity when s=256 ms=256\text{ m} . Let us find t from s=160t−16t2s=160t-16{{t}^{2}} by substituting this value of s.
⇒256=160t−16t2 ⇒16t2−160t+256=0 ⇒16(t2−10t+16)=0 ⇒t2−10t+16=0 \begin{aligned} & \Rightarrow 256=160t-16{{t}^{2}} \\\ & \Rightarrow 16{{t}^{2}}-160t+256=0 \\\ & \Rightarrow 16\left( {{t}^{2}}-10t+16 \right)=0 \\\ & \Rightarrow {{t}^{2}}-10t+16=0 \\\ \end{aligned}
Let us factorize the above polynomial by splitting the middle term.
⇒t2−2t−8t+16=0 ⇒t(t−2)−8(t−2)=0 ⇒(t−8)(t−2)=0 ⇒(t−8)=0,(t−2)=0 ⇒t=8 sec ,2 sec \begin{aligned} & \Rightarrow {{t}^{2}}-2t-8t+16=0 \\\ & \Rightarrow t\left( t-2 \right)-8\left( t-2 \right)=0 \\\ & \Rightarrow \left( t-8 \right)\left( t-2 \right)=0 \\\ & \Rightarrow \left( t-8 \right)=0,\left( t-2 \right)=0 \\\ & \Rightarrow t=8\text{ sec },2\text{ sec} \\\ \end{aligned}
Now, let us find the value of v(t)v\left( t \right) at t=8 and t=2t=8\text{ and }t=2 .
⇒v(8)=160−32×8 ⇒v(8)=160−256 ⇒v(8)=−96 m/s \begin{aligned} & \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{aligned}
Now, we have to find v(t)v\left( t \right) at t=2t=2 .
⇒v(2)=160−32×2 ⇒v(2)=160−64 ⇒v(2)=96 m/s \begin{aligned} & \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{aligned}
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.
⇒a=dvdt=d2sdt2\Rightarrow a=\dfrac{dv}{dt}=\dfrac{{{d}^{2}}s}{d{{t}^{2}}}
Students must be thorough with the formulas of derivatives.