A student is running at her top speed of 4.6 m/s to catch a bus, which is stoppe

Question

A student is running at her top speed of 4.6 m/s to catch a bus, which is stopped at the bus stop. When the student is still a distance 39.4 m from the bus, it starts to pull away, moving with a constant acceleration of 0.176 m/s^2 .

1-For how much time does the student have to run at 4.6 m/s before she overtakes the bus?

2-For what distance does the student have to run at 4.6 m/s before she overtakes the bus?

3-When she reaches the bus, how fast is the bus traveling?

4-What is the minimum speed the student must have to just catch up with the bus?

5-For what time does she have to run in that case?

6-For what distance does she have to run in that case?
















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Part B

For what distance does the student have to run at 4.6 before she overtakes the bus?

Explanation / Answer

A) = catches bus Distance the student must run = s + 38.9 = vt = 4.6 t Distance the bus moves = s = 1/2 * 0.177 * t^2 1/2 * 0.177 * t^2 + 38.9 = 4.6 t That is a quadratic 0.0885t^2 - 4.6t + 38.9 = 0

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