Question

A rocket starts from rest and moves upward from the surface of the earth. For the first 10s of its motion, the vertical acceleration of the rocket is given by ay=(3.00m/s^3)t, where the +y-direction is upward.

Answer 1

The rocket’s height at 10 seconds is 500 meters.

The speed of the rocket when it is 345 m above the surface of the earth is 117 m/s.

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Answer 2

Since the rocket’s acceleration is 3.00 m/s^3 * t, its acceleration is increasing at the rate of 3 m/s^3 each second. The equation for its velocity at a specific time is the integral of the acceleration equation.

vf = vi + 1.5 * t^2, vi = 0
vf = 1.5 * 10^2 = 150 m/s
This is the rocket’s velocity at 10 seconds. The equation for its height at specific time is the integral velocity equation

yf = yi + 0.5 * t^3, yi = 0
yf = 0.5 * 10^3 = 500 meters
This is the rocket’s height at 10 seconds.

Part B
What is the speed of the rocket when it is 345 m above the surface of the earth?
Express your answer with the appropriate units.


Use the equation above to determine the time.

345 = 0.5 * t^3
t^3 = 690
t = 690^⅓
This is approximately 8.837 seconds. Use the following equation to determine the velocity at this time.

v = 1.5 * t^2 = 1.5 * (690^⅓)^2
This is approximately 117 m/s.


The graph of height versus time is the graph of a cubic function. The graph of velocity is a parabola. The graph of acceleration versus time is line. The slope of the line is the coefficient of t. This is a very different type of problem. For the acceleration to increase, the force must be increasing. To see what this feels like slowly push the accelerator pedal of a car to the floor. Just don’t do this so long that your car is speeding!!