00:01
Welcome to this numerate tutorial.
00:03
If we start on question one of the problem set, we can define acceleration as velocity divided by time or the change in velocity over the function of time.
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So if we equal the acceleration to zero, this will translate to constant velocity divided into time.
00:31
So if we pick the correct answer, displacement of car in each one second time interval is the same.
00:48
So therefore, the correct answer is c.
00:52
As for the next problem in the question set, a would be the correct answer, and here's why.
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See, we have ground level.
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We have our object.
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We throw the object up.
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It reaches its top height of motion or its maximum height opposing the gravitational acceleration downward.
01:23
This translates to the acceleration and velocity object at the top of motion or maximum height are equal to zero.
01:34
So we have a negative 9 .81 meters per second squared acceleration constantly opposing the object downward...