Question

a toy rocket fired straight up into the air has height s(t)=224t - 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 = 10? (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? (a) the rockets initial velocity is ft/sec.

Explanation:

Step1: Find the velocity function

The velocity function $v(t)$ is the derivative of the position function $s(t)$. Given $s(t)=224t - 16t^{2}$, using the power - rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$, we have $v(t)=s^\prime(t)=\frac{d}{dt}(224t - 16t^{2})=224-32t$.

Step2: Find the initial velocity (answer for part a)

Substitute $t = 0$ into the velocity function $v(t)$. So $v(0)=224-32\times0 = 224$ ft/sec.

Step3: Find the velocity at $t = 2$ (answer for part b)

Substitute $t = 2$ into $v(t)$. Then $v(2)=224-32\times2=224 - 64=160$ ft/sec.

Step4: Find the acceleration function

The acceleration function $a(t)$ is the derivative of the velocity function. Since $v(t)=224 - 32t$, then $a(t)=v^\prime(t)=\frac{d}{dt}(224-32t)=- 32$ ft/sec². The acceleration is constant, so at $t = 10$, $a(10)=-32$ ft/sec² (answer for part c).

Step5: Find the time when the rocket hits the ground (answer for part d)

The rocket hits the ground when $s(t)=0$. So we set $224t-16t^{2}=0$. Factor out $16t$: $16t(14 - t)=0$. This gives two solutions $t = 0$ (corresponds to the launch time) and $t = 14$ seconds.

Step6: Find the velocity when the rocket hits the ground (answer for part e)

Substitute $t = 14$ into the velocity function $v(t)$. Then $v(14)=224-32\times14=224 - 448=-224$ ft/sec.

Answer:

(a) 224
(b) 160
(c) -32
(d) 14
(e) -224