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In a rocket contest, a rocket is launched from rest vertically upward from the ground. The rocket increases its velocity with a constant acceleration for 15 seconds until all fuel is used. When the rocket returns to the ground, its speed is 488 m/s.

What is the upward acceleration of the rocket due to burning fuel? Assume that the entire motion of the rocket occurs in the vertical direction and air resistance is not present.

Answer :

The upward acceleration of the rocket due to burning fuel is approximately 32.53 m/s².

To solve this problem, we can use the equations of motion for uniformly accelerated motion. We'll consider the rocket's motion in the vertical direction, assuming upward motion as positive.

Let's denote:

u = initial velocity (0 m/s as the rocket starts from rest)

v = final velocity (488 m/s)

t = time taken (15 seconds)

a = acceleration (unknown)

The equation we'll use is:

v = u + at

Substituting the given values into the equation:

488 m/s = 0 m/s + a * 15 s

We can solve this equation for the acceleration (a):

a = 488 m/s / 15 s

a ≈ 32.53 m/s²

Therefore, the upward acceleration of the rocket due to burning fuel is approximately 32.53 m/s².

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