Question

a.) The acceleration of a rocket fired vertically upwards t seconds after launch is 20+2t ms−2 (as a rocket burns fuel it becomes lighter, so accelerates more quickly). What is the second order differential equation for the height of the rocket.

h′′=  

b.) What is the general solution? (Please use A as the first constant of integration and B as the second):

General solution: h =  

c.) Use the fact that at t = 0 the rocket was on the ground and not moving to find the particular solution that gives the height of the rocket. How high was the rocket after 10 seconds? How fast was it moving then? (hint: acceleration is the rate of change of velocity. The velocity of the rocket is the rate of change of what?)

Height = _______ meters

Velocity = _______ meters/second

Answer 1

d'h alt) then 3

aun out UP on 1333 33 m 1333, 22m 3uo mls