Grade 12 Challenging Kinematics

[IQmissing]

I'm in a pickle with this question. Turning to anyone who can help me please.

A pedestrian is running at his maximum speed of 6.0 m/s trying to catch a bus that is stopped at a traffic light. When he is 16 m. from the bus, the light changes and the bus pulls away from the pedestrian with an acceleration of 1.0 m/s/s.

a) Does the pedestrian catch the bus and, if so, how far does he have to run? (If not, what is the pedestrian's distance of closest approach?)
b) How fast is the bus moving when the pedestrian catches it? ( or at the distance of closest approach)

I think I know that for the pedestrian to catch the bus, the displacements have to be the same, along with time. The pedestrian also has to make up for the 16m. so his displacement would be d+16....but I'm stuck, I don't know how to answer the question. Please Help. Thanks

[Integral]

It looks like you have the right idea for the distance. Now you just need to express the distance in terms of the knowns in your problem. You should have seen the equations you need in class or in your text.

What have you been given to work with?

[IQmissing]

Haha, the problem is that I have a similar question sitting in my locker at school, and this question is due tomorrow, I'm just so confused, because I seems like the answer is right there. I have many equations to work with, but I seem to have 2 variables for both the bus and the pedestrian. If you could suggest a way to either get the time or the displacement, I could work off of that.

[IQmissing]

pedestrian : velocity = 6.0m/s
Acceleration = 0 m/s/s (constant velocity, not sure if correct)
displacement = displacement of bus + 16m
time = same as bus

bus: acceleration = 1.9 m/s/s
V(initial) = 0m/s
V(final) = ?
displacement = ?
time = ?

[IQmissing]

The velocity (final) for the bus would have to be equal to the pedestrian (thinking out loud, stop me if I'm wrong). therefore V(f) = 6.0 m/s

You then use the equation v(f)^2 = v(i)^2 + 2ad
(6.0m/s)^2 = 2(1.0m/s/s) (d)
36 m/s = 2.0 m/s/s (d)
18 m = displacement of bus

therefore 18 + 16 = displacement of person
34 m = displacement of person.

[IQmissing]

thats definately wrong

[Päällikkö]

You did the bus part right, but the displacement of the pedestrian should have a -16, as he starts behind the bus. Why should the displacement of the person have anything to do with the velocity of the bus (your 18 + 16) ?


The problem gets really simple if you think about curves.