A rocket moves upward, starting from rest with an acceleration of

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The discussion centers on calculating the height a rocket reaches after accelerating upward at 29.0 m/s² for 8.00 seconds before running out of fuel. Participants clarify the distinction between acceleration and velocity, emphasizing that 29.0 m/s² represents acceleration, not velocity. The relevant equation for displacement during uniform acceleration is highlighted, specifically deltaX = (1/2) * acceleration * (time²). The importance of understanding these concepts is stressed to solve the problem accurately. Overall, the thread focuses on applying the correct equations of motion to determine the rocket's height.
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Homework Statement



A rocket moves upward, starting from rest with an acceleration of 29.0 m/s^2 for 8.00 s. It runs out of fuel at the end of the 8.00 s but does not stop. How high does it rise above the ground? (in meters)


Homework Equations



deltaX=(1/2)final velocity * change in time ?

The Attempt at a Solution




i really don't understand which of the equations of straight-line motion would be relevant, and how, for example: 29m/s^2 is DIFFERENT than just 29m/s
 
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Yes, 29.0 is the acceleration not velocity... what are your displacement equations for uniform acceleration?
 
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