Question
Asked Sep 13, 2019
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A rocket blasts off and moves straight upward from the launch pad with constant acceleration. After 2.8 seconds the rocket is at a height of 91 meters.

(A) What are the magnitude and direction of the rocket's acceleration?

(B) What is it's speed at this time?

Can you please solve the question using one or more of these equations:

  • v = v0 +at
  • vav = (1/2) times (v+ v)
  • x = x+ (1/2) times (v0 + v) times t
  • x = x+ v0t + (1/2)at2
  • v= v02 + 2a(x - x0)

 

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Expert Answer

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Step 1

Using the equation of motion,

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x=ar 91m-(0m/s)+(a) (2.8s) 2(91m (2.8s) 23.21m/s (directedupwards)

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Step 2

Speed at the give...

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v=% +at (0m/s)+(23.21m/s)(2.8s) 64.99m/s

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Science

Physics

Kinematics

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