Question

a helicopter is rising straight up in the air. its distance from the ground t seconds after takeoff is s(t) feet, where s(t)=2t² + 8t. (a) how long will it take for the helicopter to rise to 24 feet? (b) find the velocity and acceleration of the helicopter when it is 24 feet above the ground.

Explanation:

Part (a)

Step1: Set up the equation

We know the distance function is \( s(t) = 2t^{2}+8t \), and we want to find \( t \) when \( s(t)=24 \). So we set up the equation:
\( 2t^{2}+8t = 24 \)

Step2: Simplify the equation

Divide both sides of the equation by 2:
\( t^{2}+4t = 12 \)
Then, rearrange it to the standard quadratic form \( ax^{2}+bx + c = 0 \):
\( t^{2}+4t - 12=0 \)

Step3: Solve the quadratic equation

We can factor the quadratic equation. We need two numbers that multiply to - 12 and add up to 4. The numbers are 6 and - 2. So,
\( (t + 6)(t - 2)=0 \)
Setting each factor equal to zero gives \( t+6 = 0 \) or \( t - 2=0 \). Solving for \( t \), we get \( t=-6 \) or \( t = 2 \). Since time cannot be negative in this context, we discard \( t=-6 \).

Step1: Recall the formulas for velocity and acceleration

The velocity \( v(t) \) is the first derivative of the position function \( s(t) \), and the acceleration \( a(t) \) is the second derivative of the position function \( s(t) \) (or the first derivative of the velocity function \( v(t) \)).

Step2: Find the velocity function

Given \( s(t)=2t^{2}+8t \), we find the first derivative using the power rule \( \frac{d}{dt}(t^{n})=nt^{n - 1} \).
\( v(t)=s^\prime(t)=\frac{d}{dt}(2t^{2}+8t)=2\times2t+8=4t + 8 \)

Step3: Find the acceleration function

We find the derivative of the velocity function \( v(t) = 4t+8 \).
\( a(t)=v^\prime(t)=\frac{d}{dt}(4t + 8)=4 \)

Step4: Evaluate velocity and acceleration at \( t = 2 \) (from part (a))

First, evaluate the velocity at \( t = 2 \):
\( v(2)=4\times2+8=8 + 8=16 \) feet per second
The acceleration function \( a(t) = 4 \) is a constant, so at \( t = 2 \), \( a(2)=4 \) feet per second squared.

Answer:

\( t = 2 \) seconds

Part (b)