phys 16 qestion

Question

A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 ss of its motion, the vertical acceleration of the rocket is given by ay=(2.70m/s3)tay=(2.70m/s3)t, where the +y+y-direction is upward.

What is the height of the rocket above the surface of the earth at ttt = 10.0 ss?

Answer 1

We will apply Given, Asked, Picture, and Solve strategy GAPS

Given: Initial velocity Vy0 = 0, Initial Height y = 0 and acceleration Ay = 2.7t

Asked: Height of rocket at t = 10 sec

Picture: Rocket motion upward with variable acceleration.

Solve: Integrate acceleration twice with respect to time.

Vy = ∫ 2.7t dt = 2.7t^2/2 + C, C = 0 initial velocity = 0 m/s

Y = ∫ 2.7t^2/2 dt = 2.7t^3/6 + C, C = 0 Initial height = 0 m

Height at t = 10 s, substitute t = 10 in expression of Y

Y = 2.7 t^3/6 = 2.7 x 10^3/6 = 450 m

I hope this helps.

Shailesh (Sky) Kadakia, Expert Physics and Math tutor.

with WYZANT, Inc.