Homework Help: Constant Acceleration Problem - Two bodies with different accelerations

1. Sep 10, 2011

Nickg140143

1. The problem statement, all variables and given/known data
At the instant the traffic light turns greeen, Stan Speedy starts from rest and accelerates at "a". At the same instant Kathy Kool starts from rest "d" behind Stan and accelerates at "5/4a" (not sure, but I think this is meant to be (5/4)a or 5a/4, my attempts with "5/4a" gave me strange answers).

Given [a, d], determine:
a.How far beyond the starting point do they meet again.
b.The time before they meet again.
c.How fast is each moving when they meet again.

2. Relevant equations
displacement & time:
x=x0+v0t+(1/2)at2

velocity & time:
v=v0+at

velocity & displacement:
v2=v20+2aΔx

3. The attempt at a solution
I've attached an image that shows a diagram I drew to help me solve the problem. Whether it be correct or not, that I do not know.

The only approach I could think of was to first set x=x0+v0t+(1/2) for each car equal to each other, that is, when each car would reach x1

Since each car is starting from rest, I believe that v0=0

Stan's Car:
x1=(0)t+$\frac{1}{2}$at2
x1=$\frac{1}{2}$at2

Kathy's Car:
x1=(0)t+$\frac{1}{2}$$\frac{5a}{4}$t2
x1=$\frac{5a}{8}$t2

Stan's x1 = Kathy's x1
$\frac{1}{2}$at2=$\frac{5a}{8}$t2

My reasoning for this was to then solve for t so that I would then know at what time that these two cars would be at the same point, and then I could use t to help me calculate the other unknowns, but whenever I try to solve for t, I get 0, which doesn't make much sense to me.

I think my problem is that I'm not accounting for the starting distance between Kathy's and Stan's car at the beginning of the problem.

I've been racking my head over this for hours
(physics and math aren't necessarily by best subjects )

A great many thanks to any who can point me in the right direction on this.

Attached Files:

• phy4AH2P4.png
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2. Sep 11, 2011

ehild

Your picture is very nice and you correctly showed that Stan's initial position is X(0)=d if Kathy is at the origin initially, x(0)=0. Put this fact into the equation of Stan's position. The equation you cited is x=x(0) +v(0)t +(a/2)t^2.

By the way, 5/4a = 5/4*a=(5/4)*a. Nothing means multiplication. If you have multiplication and division, perform them from right to left. If a=4, it is 4 and not 5/16.

ehild

3. Sep 11, 2011

Nickg140143

*Face Palm*

After taking another look at the equation for Stan's position, I noticed the equation I wrote in my notes was x=v(0)t +(a/2)t^2, not x=x(0) +v(0)t +(a/2)t^2.
(you can even see it in my drawings in the time left corner)

After fixing that and following your suggestion, I was able to get some results that seem promising. Once I fix my notes, I'll try to run through this problem in its entirety and see if I'm successful.

Thanks a lot Ehild, I'm glad I decided to post this on the forums.

4. Sep 11, 2011

HallsofIvy

But, mathematically, it would not be unreasonable to interpret "5/4a" as "5/(4a)". I believe that was what Nickg140143 was referring to. Of course, because a is acceleration, and so has units of "m/s^2" associated with it, "5/(4a)" would have the wrong units for an acceleration, "s^2/m".

5. Sep 11, 2011

Nickg140143

Yes, the way that part of the problem was typed onto our assignment sheet confused me when I first looked at it, so aside from the units being strange if I interpreted it at 5/(4a), It made more sense to me that the acceleration of Kathy is 5/4 of "a" [(5/4)(a) or (5a)/4], which is slightly more than Stan's acceleration "a", since she eventually catches up with him.