A UCI student is running at her top speed of 5.4 m/s to catch a Shuttle bus, whi

Question

A UCI student is running at her top speed of 5.4 m/s to catch a Shuttle bus, which is stopped at the bus stop. When the student is still a distance 40.5 m from the Shuttle bus, it starts to pull away, moving with a constant acceleration of 0.177 m/s2 .

Part A

For how much time does the student have to run at 5.4 m/s before she overtakes the bus?

Part B

For what distance does the student have to run at 5.4 m/s before she overtakes the bus?

Part C

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

Part D

If the student's top speed is 3.00 m/s , will she catch the bus?

Part E

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

Part F

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

Part G

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

Explanation / Answer

student: s = 5.4m/s * t
bus: S = 40.5m + ½ * 0.177m/s² * t²
When does s = S?
5.4t = 40.5+0.0885t²
quadratic in t, solutions at
a)t = 8.75s, 52s (the second one is when the bus catches back up to the student!)

b)s = 5.4m/s * 8.75s = 47.5 m

c) Vbus = at = 0.177m/s² * 8.75s = 1.548 m/s

d) In this case the quadratic equation would not be
0 = 0.885t² - 5.4t - 40.5, but
0 = 0.885t2 - 3t - 40.5

t=16.94

So she would still catch the bus.

e)we used
0.0885t² - 5.4t + 40.5 = 0 to solve a thru c.
t = (-b ± (b² - 4ac) ) / 2a
I think that at the minimum speed, that the discriminant will have a value of 0 so that t will have only one root (and not two). So b² = 4ac = 4(0.0885)(40.5) = 14.337 and b = 3.78 m/s.

f) Check:
3.78t = 40.5 + 0.0885t²
Single root at t = 22.235 s e
(more precise value was v = 3.581165 m/s)

g) s = 3.78m/s * 22.235s = 84.049 m

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