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Question

The Space Shuttle on reentry, slows its velocity from 565km/hr to 496km/hr in 15 seconds, then in the next 17 seconds it slows to 346km/hr at touch down. What is its acceleration during this time frame.

Answer 1

I'm not exactly sure what 'time frame' this question is asking for (that is if it's asking for the entire reentry process). I'll show you how to get the two intervals' acceleration, and if it's asking for the accel of the entire process, you can use my method to find the answer. ^_^

Main equations here:
a (acceleration) = change in velocity / change in time
change in velocity = final velocity - initial velocity
change in time = final velocity - initial velocity


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The first acceleration is during the initial reentry.

The equation for finding acceleration requires finding the change in velocity and also the change in time. So do that first.

change in velocity = final velocity - initial velocity
change in velocity = (496km/hr) - (565km/hr)
change in velocity = - 69 km/hr

change in time = final time - initial time
change in time = (15 seconds) - (0 seconds)
change in time = 15 seconds
15 seconds = 0.25 hours

a = change in velocity / change in time
a = (-69 km/hr)/(0.25 hr)
a = - 276 km/hr2 (kilometers per hour squared)

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The second interval:

change in velocity = final - initial
change in velocity = (346km/hr) - (496km/hr)
change in velocity = - 150 km/hr

change in time = final - initial
change in time = (17.0 seconds) - (0 seconds)
change in time = 17.0 seconds
17.0 seconds = 0.283 hr

a = change in velocity / change in time
a = (-150 km/hr) / (0.283 hr)
a = -530 km/hr2 (kilometer per hour squared)

Answer 2

Average acceleration (a) is simply the change in velocity (del v) divided by the time interval (del t) that change took place in. Thus, a = del v/del t = (496 - 565)/15 for the first increment of deceleration (the answer has a minus sign, which indicates deceleration). And a' = del v'/del t' = (346 - 496)/17 for the next increment. You can do the work.

If by "time frame" you mean the entire 32 seconds of deceleration, you can use the same a = del v/del t, but use del v = 346 - 565 and del t = 32 instead.