A student is running at her top speed of 5.5 to catch a Shuttle bus, which is st

Question

A student is running at her top speed of 5.5 to catch a Shuttle bus, which is stopped at the bus stop. When the student is still a distance 39.2 from the Shuttle bus, it starts to pull away, moving with a constant acceleration of 0.163 .

For how much time does the student have to run at 5.5 before she overtakes the bus?

For what distance does the student have to run at 5.5 before she overtakes the bus?

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

If the student's top speed is 2.80 , will she catch the bus?

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

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

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

Explanation / Answer

The speed of the student vs = 5.5m/s

the distance between bus and student d = 39.2m

the acceleration of the bus a = 0.163 m/s^2

now the distance travled by bus during the time student traveld

         s1 = 1/2at^2 = (0.5)(0.163)t^2

                             = 0.0815t^2

now the total distance traveled by student

         s = s1 + d

       vs*t = 0.0815t^2 + 39.2

      5.5t = 0.0815t^2 + 39.2

     0.0815t^2 - 5.5t + 39.2=0

by solving for t = 8sec

the distance traveld d = 5.5*8 = 44m

the speed of the bus v = at   = 0.163*8 = 1.3m/s

If the speed is 2.8m/s

    2.8t = 0.0815t^2 + 39.2

by solving t = undefine

it means that she unable to catch the bus

The minimum speed

vs = 0.0815t + 39.2/t

dvs/dt = 0.0815 - 39.2/t^2

0 = 0.0815 -39.2/t^2

    t^2 = 39.2/0.0815

    t = 21.9 = 22s

therefore the minimum speed

     vmin = 0.0815(22) + 39.2/(22)

              = 3.57m/s

therefore the distance

    d = 3.57*22 = 78.54 m

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