A UCI student is running at her top speed of 5.1 m/s to catch a Shuttle bus, which is stopped at the bus stop. When the student is still a distance 42.0 m from the Shuttle bus, it starts to pull away, moving with a constant acceleration of 0.163 m/s2 .For how much time does the student have to run at 5.1 m/s before she overtakes the bus?

Question
Asked Oct 22, 2019
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A UCI student is running at her top speed of 5.1 m/s to catch a Shuttle bus, which is stopped at the bus stop. When the student is still a distance 42.0 m from the Shuttle bus, it starts to pull away, moving with a constant acceleration of 0.163 m/s2 .For how much time does the student have to run at 5.1 m/s before she overtakes the bus?

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Expert Answer

Step 1

Let us say that the student overtakes the bus after ‘t’ seconds.

Now the distance travelled by the bus in t seconds will be:

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at2 bus 2 We know. 0, and a = 0.163 u, 0.163t -- Eq bus 2

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Step 2

In the same time, the student travels ‘s’ meters:

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s 42 S Also s 5.1 t Thus =5.1t -42 Sbus --- Eq2

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Step 3

From Eq1 and Eq2...

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0.163 5.1 42 2 or, 62.576t -515.337 = t2 or,t-62.576t +515.337 = 0 Solving the quadratic equation,we get t -9.756s andt = 52.82s

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