Acceleration problem-how to incorporate head starts?

[sean-820]

Homework Statement

A pedestrian is running at v=6m/s (continuously) to catch a bus that is leaving from 16m in front of him. The bus pulls away and accelerates (from rest) at 1m/s/s.

How far does the pedestrian have to run to catch the bus (if at all)?

V pedestrian=6m/2
acceleration bus= 1m/s/s
V1bus= 0m/s

The Attempt at a Solution

How would i go about solving this one? Its easy to solve for there intersection, but I'm not sure how to incorporate the 16m head start by the bus into a equation.

i used: v1(delta-t)+0.5a(delta-t)^2=v(delta-t)
to find the distance at which they meet, but would the equation with the 16m be:
v1(delta-t)+0.5a(delta-t)^2=v(delta-t)+16 ?



This way doesn't look right to me so I am wondering how it should be incorporated.

So how can i incorporate the 16m head start by the bus into the equation?

Homework Equations

delta distance=v1 (delta time)+0.5(acceleration)(delta time)^2

v1 (delta time)+0.5(acceleration)(delta time)^2=v(delta time)

Times cancel leaving v1 =0 so its gone

0.5(acceleration)(delta time)=v
then

(delta time)=(2v)/a

 

[Redbelly98]

Yes, those problems where two objects start out at different places or at different times can be tricky.

Let the initial displacements of the person and bus be 0 and 16 m, respectively.

Any expression for the person's displacement should give 0 m at t=0. Likewise, any expression for the bus's displacement should give x=16m at t=0.

In this way, you can check your expressions for the displacements of the person and bus, and see if they are correct.

 

[sean-820]

Thanks for the reply

Are you suggesting this then: (v1(delta-t)+0.5a(delta-t)^2)+16=v(delta-t) ?

This was another way i was thinking and after you said that it seems like a more logical equation as the bus would be 16m ahead which would mean the time would need to be grater to also overcome this distance.

 

[Redbelly98]

Are you suggesting this then: (v1(delta-t)+0.5a(delta-t)^2)+16=v(delta-t) ?

Yes, that looks good

 

[sean-820]

ok, thanks alotone last thing, i just calculated it out and got the intersection time to be -4 seconds which I am assuming means the person missed the bus(and could of ran to catch it if he was 4 seconds earlier?) . To find how much (distance) he missed the bus by could i use d=vt

part b asks How fast is the bus moving when the pedestrian catches it/ the distance of closest approach.

 

[sean-820]

This head start is really complicating the problem for me.

How do i find how fast the bus was moving when the pedestrian was 24m behind it?


Could i do a similar thing as before and do:

((v2square-v1square)/2a)+16=v2square-v1square)/2a


I think need something like this as i don't know much about time (other then the bus was missed by 4 seconds)

To find the bus' v2 i need either to find the time it happened or that distance, but i can't seem to figure out how to isolate one of these variables while keeping the 16m head start accounted for.

 

[Redbelly98]

Let's try to get this part right first:

i just calculated it out and got the intersection time to be -4 seconds ...

I get something different. Can you show your calculation please? Using the equation you had:

(v1(delta-t)+0.5a(delta-t)^2)+16=v(delta-t)

 

[sean-820]

Displacement bus=displacement person

v1(t)+0.5a(t)^2+16=V(T)
0.5a(t)+16=V Simplify by removing the (T), -bus v1=0
0.5(1)(t)+16=(6)
(t)=(6-16)/0.5
t=-20 seconds?Did you get t=-20 seconds?
Is this what you got? I forgot to divide by 0.5. I had divide by (a) which was 1

I had t=(2v/a)-16
when i think it should of been
t=2((v/a)-16)

 

[Redbelly98]

I get positive values for t.

Displacement bus=displacement person

v1(t)+0.5a(t)^2+16=V(T)
0.5a(t)+16=V Simplify by removing the (T), -bus v1=0

If you are dividing by t, then this should be
0.5at + 16/t = V
Note the "16/t" term. Since this step does not really simplify the problem, go back to the step before. It's a quadratic equation in t, which is solvable using methods learned algebra (i.e. factoring or quadratic formula).

 

[sean-820]

Using the quadratic formula, i got x=4 and x=8 assuming I am right this time

Im hoping this is right, but I am not exactly sure what this is graphing. Right now i have a parabola with the
vertex form equation 0.5(t-6)^2-2=0
standard form 0.5(t)^2-6t+16=0

So what would this even be graphing?


Thanks for all your help btw

 

[Redbelly98]

Using the quadratic formula, i got x=4 and x=8 assuming I am right this time

Good, I got the same values

Edit:
Actually it's t=4 and t=8 ... right?

Im hoping this is right, but I am not exactly sure what this is graphing. Right now i have a parabola with the
vertex form equation 0.5(t-6)^2-2=0
standard form 0.5(t)^2-6t+16=0

So what would this even be graphing?

It's a graph of (x_bus - x_person), i.e. the displacement of the bus relative to the person.

• When the graph is positive, the bus is ahead of the person
• When the graph is negative, the bus is behind the person
• When the graph is zero, the bus and person are at the same position, which is what you were trying to find.

It might be more instructive to graph two functions on the same graph:

0.5(t)^2 + 16​

and

6t​

for the displacements of the bus and person, respectively.

Thanks for all your help btw

You're welcome.

Edit:
Whoops, still a couple of loose ends to tie up:
1. Which solution, t=4 or t=8, or both, is the correct answer?
2. How far does the pedestrian have to run to catch the bus?

 

[sean-820]

Edit:
Whoops, still a couple of loose ends to tie up:
1. Which solution, t=4 or t=8, or both, is the correct answer? t=4s
2. How far does the pedestrian have to run to catch the bus? 24m

Buses velocity when caught is 4m/s. (v2=v1+a(t))

Would t=8 or x=8 matter since the x-axis is t? Or is it just better format to use t?

Im not sure what you mean by the graph being negative/positive. Its a parabola so it goes into negative x values, but those don't matter as its time. When i graphed both the interception point was (4,24).

So I'm assuming the pedestrian catches the bus at 4. 4-8 seconds the pedestrian would be ahead of the bus. then at 8 seconds the bus would pull away and the pedestrian can't catch it again.

 

[Redbelly98]

Edit:
Whoops, still a couple of loose ends to tie up:
1. Which solution, t=4 or t=8, or both, is the correct answer? t=4s
2. How far does the pedestrian have to run to catch the bus? 24m

Buses velocity when caught is 4m/s. (v2=v1+a(t))

Yes, looks good. Problem solved!

Would t=8 or x=8 matter since the x-axis is t? Or is it just better format to use t?

Since we started out using t for time, it really should be t. Often x is used for displacement or distance (and gets plotted on the y-axis!), so that is a source of confusion, or possibly points deduction on an exam. I know algebra classes always used x for the horizontal axis, and it can take some getting used to for that to be different.

Im not sure what you mean by the graph being negative/positive. Its a parabola so it goes into negative x values, but those don't matter as its time. When i graphed both the interception point was (4,24).

Yes, negative time doesn't matter. But I meant when the "y values" are negative, between t=4 and t=8.

So I'm assuming the pedestrian catches the bus at 4. 4-8 seconds the pedestrian would be ahead of the bus. then at 8 seconds the bus would pull away and the pedestrian can't catch it again.

Yes, exactly.

 

[sean-820]

thanks so much for your help and taking the time to clarify, it was appreciated and now i feel i could answer more questions like this easily with no help