Arocket 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.

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!!

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. 

 

I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.