A toy rocket is fired straight up into the air. Let s(t) = -6t2 + 72t denote its position in feet after t seconds.

Question

(a) Find the velocity after t seconds

(b) Find the acceleration after t seconds

(c) When does the rocket reach its maximum height?

(d) What is the maximum height reached by the rocket?

Answer 1

The position equation is given by:

s(t) = -6t² + 72t

The velocity is the rate of change of the position; therefore, velocity is the derivative of the position equation:

1. s(t) = -6t² + 72t
2. v(t) = (-6)(2)t + 72
3. v(t) = -12t + 72

The acceleration is the rate of change of velocity; therefore, acceleration is the derivative of the velocity equation and acceleration is also the second derivative of the position equation:

1. v(t) = -12t + 72
2. a(t) = -12

The rocket reaches its maximum height when it stops moving upwards; therefore, using the velocity equation, and setting the velocity equation to zero (implying that the rocket is no longer moving upwards), and solving for time (t), we are able to find the amount of time it took the rocket to reach the maximum height:

1. -12t + 72 = v(t)
2. -12t + 72 = 0
3. -12t = -72
4. t = 6 s.

Now that we know how long it took for the rocket to reach the maximum height, we may use this information (t = 6 s) to find the height. We do this by taking this value, which we just found, and plugging it into the position equation:

1. s(t) = -6t² + 72t
2. s(t) = -6 (6)² + 72 (6)
3. s(t) = -216 + 432
4. s(t) = 216 units

The maximum height reached by the rocket is 216 units.