Kinematics Problem about catching up to a bus

A student is running at her top speed of 4.6 m/s to catch a bus, which is stopped at the bus stop. When the student is still a distance 38.9 m from the bus, it starts to pull away, moving with a constant acceleration of 0.177 m/s^2.

A) For how much time does the student have to run at 4.6 m/s before she overtakes the bus?

I got 10.6s for this part by setting the distance equations for each to one another and solving for time.

Then I'm not sure how to set up the rest...

B) For what distance does the student have to run at 4.6 m/s before she overtakes the bus?

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

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

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

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

Re: Kinematics Problem about catching up to a bus

Quote:

Originally Posted by

**ayyisei** A student is running at her top speed of 4.6 m/s to catch a bus, which is stopped at the bus stop. When the student is still a distance 38.9 m from the bus, it starts to pull away, moving with a constant acceleration of 0.177 m/s^2.

A) For how much time does the student have to run at 4.6 m/s before she overtakes the bus?

I got 10.6s for this part by setting the distance equations for each to one another and solving for time.

Then I'm not sure how to set up the rest...

B) For what distance does the student have to run at 4.6 m/s before she overtakes the bus?

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

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

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

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

you found t = 10.6 sec ...

b) $\displaystyle d_s = v_s \cdot t$

c) $\displaystyle v_b = a_b \cdot t$

for parts d) , e), and f) , you'll need to find the student velocity $\displaystyle v_s$ such that the graph of the student's position line is tangent to the bus' position curve. in other words, the student and bus are at the same position and both are traveling the same speed. therefore, solve the simultaneous system ...

$\displaystyle v_s \cdot t = 38.9 + \frac{1}{2}a_b t^2$ and $\displaystyle v_s = v_b = a_b \cdot t$

Re: Kinematics Problem about catching up to a bus

For parts D-F, can I solve for v_s first then use that to solve for the next equation?

Re: Kinematics Problem about catching up to a bus

Quote:

Originally Posted by

**skeeter** $\displaystyle v_s \cdot t = 38.9 + \frac{1}{2}a_b t^2$ and $\displaystyle v_s = v_b = a_b \cdot t$

Is the second equation supposed to be like that or is it:

v_s - v_b = a_b * t

Re: Kinematics Problem about catching up to a bus

Quote:

Originally Posted by

**ayyisei** Is the second equation supposed to be like that or is it:

v_s - v_b = a_b * t

as i posted it ... I told you that the student's speed had to equal the bus' speed if he/she were to just catch it.

Re: Kinematics Problem about catching up to a bus

Quote:

Originally Posted by

**skeeter** as i posted it ... I told you that the student's speed had to equal the bus' speed if he/she were to just catch it.

So the t is t = 10.6 s that I found.

The v_s is the constant velocity that the student is running at (4.6 m/s).

The v_b is the velocity of the bus and the a_b its acceleration.

I think I'll have to substitute v_s from the first equation into the second one?

Re: Kinematics Problem about catching up to a bus

Quote:

Originally Posted by

**ayyisei** So the t is t = 10.6 s that I found. incorrect ... it will be different because you're trying to find the **minimum speed** the student can run and just catch the bus.

...

Re: Kinematics Problem about catching up to a bus

Oh okay. I understand what to do now. Thank you very much!