# 1 Dimensional Kinematics Problem

1. Sep 9, 2010

### Skizye

Hi, I've been stuck on this problem for a bit and I am not really sure how to go about solving it.

1. The problem statement, all variables and given/known data

As a traffic light turns green, a waiting car starts with a constant acceleration of 6.0 m/s². At the instant the car begins to accelerate, a truck with a constant velocity of 21 m/s passes in the next lane.
a. How far will the car travel before it overtakes the truck?
b. How fast will the car be traveling when it overtakes the truck?

2. Relevant equations

So far, all we have been working with in class are these:

$$v = v_0 + a t$$
$$v^2 = v_0^2 + 2 a (x - x_0)$$
$$x = x_0 + v_0 t + (1/2)a t^2$$

3. The attempt at a solution
I've tried several different angles for this problem, but so far I haven't been able to solve for any of the variables besides the ones given in the problem.
I set up my variable table

Car
v =
$$v_0 = 0 m/s$$
$$x - x_0 = ___$$
a = 6.0 m/s²
t =

Truck
v = 21 m/s
$$v_0 = 21 m/s$$
$$x - x_0 = ___$$
a = 0 m/s
t =

I tried guessing and checking numbers for t but that is probably not the cleanest way to solve this problem, if it's even possible to do at all. I'm not really sure how to solve for any of the variables here, any help would be appreciated, thanks!

2. Sep 9, 2010

### collinsmark

Hello Skizye,

Welcome to Physics Forums!

You need to show your work before we can help you. But if you absolutely can't get started, here is a quick hint: At the moment the car overtakes the truck, both the car and the truck have traveled an equal distance (as in "=" distance ).

3. Sep 9, 2010

### Skizye

Thanks, collinsmark! I figured it out with your hint by plugging in different times to the formula until their distances were equal. I'll be sure and post all of my work next time! =]

4. Sep 9, 2010

### collinsmark

Okay, but I'm not sure I agree with your approach. http://www.websmileys.com/sm/sad/021.gif. Such physics problems are not supposed to be solved iteratively.

What I was hinting at was something like this (this is one of a few ways to solve this problem, btw):

(i) Find an equation for the truck's distance, as a function of time t.
(ii) Find an equation for the car's distance, as a function of time t.
(iii) Set the above two equations equal to each other and solve for t. (You can do this because at the moment that the car overtakes the truck, the distances are equal. You are solving for the specific time that the distances are equal.) [Edit: But don't do this step iteratively, instead use algebra!]

(iv) Plop that t value you just found into the equation for the car's distance, and solve for the car's distance. That's the answer to part a.
(v) To double check your answer, substitute that same t value into into the truck's equation for distance, and calculate the distance. If things work out, both distances will be the same.

Last edited: Sep 9, 2010
5. Sep 9, 2010

### Skizye

That's a lot easier and cleaner than what I did.
I wound up with a time of 7 s, distance of 147 m, and velocity of 42 m/s. Does that sound about right?

6. Sep 9, 2010

### collinsmark

'Looks good to me!