# How far does the student run before catching the bus?

## Homework Statement

A student is trying to catch a bus that is stopped at a red light. She runs toward it at a constant speed of 6.0m/s. When she is 16m away from the bus, the light turns green and the bus pulls away with an acceleration of 1.0m/s^2.
(a) How far does the student run before catching the bus?
(b) How fast is the bus travelling when the student catches up with it?

## The Attempt at a Solution

The main problem i'm having with this questions is getting a grip on uniform acceleration. The questions says that the bus moves with an acceleration of 1.0 m/s^2, so does that mean every second the bus moves 1.0m, or every second the bus moves 1.0m more than it did the previous second? So at 1s it moves 1m, at 2s it moves 3m, at 3s it moves 4m etc.

## Answers and Replies

SammyS
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It's the second one: every second the bus moves 1.0m more than it did the previous second.

However, at the beginning of the first second the velocity of the bus is zero. At the end of the first second, the velocity of the bus is 1 m/s. So during the first second the average velocity of the bus is 1/2 m/s .

Do you have any kinematic equations?

Vf^2= Vi^2 +2a(delta)d
(delta)d= Vf(delta)t - 1/2*2(delta t)^2
I'm just really confused at this point and have no idea where to start.

SammyS
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Δd= Vi(Δt) + (1/2)(a)(Δt)2 looks like what you need, but you need one version for the student and one version for the bus. (Somehow, you have an extra 2 in your equation & a few other different things. sign & final rather than initial velocity.)

The student runs at a constant velocity of 6.0 m/s. What's her acceleration?
Also, she is she is 16 m from the bus when it begins to move. How much greater must her displacement be than that of the bus, for her to catch the bus?

The bus starts from rest. What is its initial velocity?

Her acceleration would be 0m/s^2 because she is moving at a constant velocity.
The bus's initial velocity would be 0m/s since starting from rest.
After 2.66s the student would have run 16m
At this point the bus would have already travelled 3.5m
Δd= Vi(Δt) + (1/2)(a)(Δt)2
= (1/2)(a)(Δt)2
=(0.5)(1.0m/s^2)(2.66s)^2
=3.5m

I need to find where the two displacement values overlap, yes?

For the girl:
Δd=4.0s (6.0m/s)
Δd= 24m

Δd= Vi(Δt) + (1/2)(a)(Δt)2
= (1/2)(a)(Δt)2
=(0.5)(1.0m/s^2)(4.0s)^2
=8m
8m+16m
Δd=24m

Therefore, the student has to run 24m before catching the bus.

Maybe I over reading this, but for a), it asked how how far does she have to run and the problem says 16m...

How far does she have to run before catching the bus, but the bus begins to move as she begins to run after it.

SammyS
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I see that you have the right answer, but it wasn't obtained in a very general way.

For bus:
Δd = (1/2)(1.0)(Δt)2

For student:
Δd is 16 meters more than the above, so Δd = (1/2)(1.0)(Δt)2 + 16

But also, Δd is also given by Δd = (6.0) (Δt)

Equating these two gives: (1/2)(1.0)(Δt)2 + 16 = (6.0) (Δt) .

Solve this for Δt. You'll get a quadratic equation with two solutions. Only the positive one makes sense.​

why does Δd = (1/2)(1.0)(Δt)2 + 16
If the student is running at a constant speed her acceleration would be 0m/s^2 not 1.0m/s^2 would it not?

SammyS
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Because (1/2)(1.0)(Δt)2 is the bus's displacement and hers is 16 m more than that of the bus giving (1/2)(1.0)(Δt)2 + 16 .