Question

a toy rocket fired straight up into the air has height s(t)=112t - 16t² feet after t seconds.
(a) what is the rockets initial velocity (when t = 0)?
(b) what is the velocity after 2 seconds?
(c) what is the acceleration when t = 5?
(d) at what time will the rocket hit the ground?
(e) at what velocity will the rocket be traveling just as it smashes into the ground?
(c) when t = 5, the acceleration of the rocket is - 32 ft/sec².
(d) the rocket will hit the ground seconds after it is launched.
(e) when the rocket smashes into the ground, it will be traveling at ft/sec.

Explanation:

Step1: Recall velocity - derivative of position

The velocity function $v(t)$ is the derivative of the position function $s(t)$. Given $s(t)=112t - 16t^{2}$, using the power - rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$, we have $v(t)=s^\prime(t)=112-32t$.

Step2: Recall acceleration - derivative of velocity

The acceleration function $a(t)$ is the derivative of the velocity function. Since $v(t)=112 - 32t$, then $a(t)=v^\prime(t)=-32$. The acceleration is constant and does not depend on $t$.

Step3: Find when the rocket hits the ground

The rocket hits the ground when $s(t)=0$. So we set $112t-16t^{2}=0$. Factor out $16t$: $16t(7 - t)=0$. This gives two solutions $t = 0$ (corresponds to the launch time) and $t = 7$ seconds.

Step4: Find the velocity when it hits the ground

We know $v(t)=112-32t$. Substitute $t = 7$ into the velocity function: $v(7)=112-32\times7=112 - 224=-112$ ft/sec.

Answer:

(d) 7
(e) - 112