Question

a toy rocket fired straight up into the air has height ( s(t) = 144t - 16t^2 ) feet after ( t ) seconds. (a) what is the rocket’s initial velocity (when ( t = 0 ))? (b) what is the velocity after 3 seconds? (c) what is the acceleration when ( t = 8 )? (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?

Explanation:

Part (a)

Step1: Recall velocity as derivative of position

The velocity function \( v(t) \) is the derivative of the position function \( s(t) \). Given \( s(t) = 144t - 16t^2 \), we find the derivative using the power rule. The derivative of \( t^n \) is \( nt^{n - 1} \).
So, \( v(t)=\frac{d}{dt}(144t - 16t^2)=144 - 32t \).

Step2: Evaluate velocity at \( t = 0 \)

Substitute \( t = 0 \) into \( v(t) \): \( v(0)=144 - 32(0)=144 \).

Step1: Use the velocity function from part (a)

We have \( v(t)=144 - 32t \).

Step2: Evaluate at \( t = 3 \)

Substitute \( t = 3 \) into \( v(t) \): \( v(3)=144 - 32(3)=144 - 96 = 48 \).

Step1: Recall acceleration as derivative of velocity

The acceleration function \( a(t) \) is the derivative of the velocity function \( v(t) \). We have \( v(t)=144 - 32t \), so the derivative is \( a(t)=\frac{d}{dt}(144 - 32t)=- 32 \).

Step2: Evaluate at \( t = 8 \)

Since the acceleration function is constant (\( a(t)=-32 \) for all \( t \)), when \( t = 8 \), \( a(8)=-32 \).

Answer:

The initial velocity is \( \boldsymbol{144} \) feet per second.

Part (b)