A student is running at her top speed of 5.4m/s to catch a bus, which is stopped

Question

A student is running at her top speed of 5.4m/s to catch a bus, which is stopped at the bus stop. When the student is still a distance 41.3m from the bus, it starts to pull away, moving with a constant acceleration of 0.173m/s2 .

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

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

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

d) If the student's top speed is 2.70m/s , will she catch the bus?

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

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

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

Explanation / Answer

a. apply S = 0.5 at^2

so S = 0.5 * 0.173 * t^2

S = 0.0865 t^2

also as Distance d = vt

d = 5.4t

but to catch the bus, he travelled as distance of 41.3 m farther than bus

so

d = 5.4t + 41.3

so time taken to catch the bus is 0.0865t^2   = 5.4t + 41.3

or   0.0865 t^2 -5.4t -41.3 = 0

solving this

t = -(-5.4)

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