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If The last frame before the shuttle begins to move is 140 and the shuttle travels 56 meters in 243 frames. (a) If each frame is 24 seconds what is the time elapsed?(b) Assuming constant acceleration, at what rate is the shuttle accelerating?(c) If the shuttle continued to move with thisacceleration, what speed would it reach 76s after launch?(d) If the shuttle traveled directly upwards, what would its altitude be at 76 s?

Respuesta :

Part a)

The number of frames between 140 and 243 is 103. Since each frame is equivalent to 24 seconds, multiply 103 by 24s to find the time elapsed:

[tex]t=103\times24s=2472s[/tex]

Part b)

Assuming constant acceleration, a the distance d traveled by an object starting at rest during a time t is:

[tex]d=\frac{1}{2}at^2[/tex]

Isolate a from the equation and replace d=56m and t=2472s to find the acceleration the shuttle:

[tex][/tex]

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