# Having trouble finding equation of one question.

1. May 11, 2014

### Sullivan

Have a very basic physics assignment due tomorrow, and am having trouble finding out an equation for one question.

Here is the problem. I'm sorry it's so basic. I've never done physics in my life. I find it beautiful and fascinating, but I am just starting out. I'd love if someone could give me a hand. I have a hard time sometimes deciphering word problems like these:

A pickup truck, traveling north at 144 km/h, overtakes and passes a police car traveling north at 28 m/s. Just as the pickup passes the police car, the car accelerates at 2.0 m/s2 until it catches up with the pickup.

Assuming the pickup does not speed up or slow down, calculate:
a) the time elapsed until the police car catches the pickup
b) the displacement of the police car and the pickup.
c) the final velocity of the police car

So we know the velocity of the pickup (p) to be 40 m/s, it's acceleration is 0, as it's velocity is constant, we don't know it's time or displacement.

The car's (c) initial velocity is 28 m/s, and it's final velocity is unknown. It's acceleration is 2.0 m/s2 and it's displacement is also unknown.

I've set t(p) = t(c), but every calculation I get has too many unknown variables, so I'm thinking substitution will be necessary. I just can't for the life of me figure out what that substition is. I'd really love some help on this. Thanks so much! I'll continue to work on this myself in the meantime. I thought maybe I could find time(p) by using the a=v/t equation, but since a=0, I get a division by zero.

2. May 11, 2014

### jackarms

Hey there, and welcome to Physics Forums.

To start, don't get discouraged about having trouble with a problem, even if it seems simple. Physics requires a great deal of problem solving, and anyone who is good at physics has had to do a ton of problems and exercises with it.

For the first part, try setting up equations for the displacement of each of the vehicles. You know that the car eventually catches up to the truck, so the displacements must be equal. If you set these equal to each other, you can solve for the time, and you shouldn't be stuck with other unknowns.

Using your method of t(p) = t(c) should also work, but I think the problem is going to be solving for t in these equations. Especially if the time is squared, it's going to get complicated, so I would stick to the basic forms.

3. May 11, 2014

### Sullivan

Thank you very much for the warm welcome! I'm dedicated to getting an A in this course, so your help is really appreciated. I try my best to figure things out on my own but sometimes I have to ask for help.

So... If I understand correctly, since the displacement and time will be the same, d(p) = d (c), v initial (p) * t + 1/2 * a (p) * t^2 = v initial (p) * t + 1/2 * a (c) * t^2
so 40 m/s * t + 1/2 * 0 m/s^2 * t^2 = 28 m/s + 1/2 * 2.0 m/s2 * t^2
12 m/s = 1 m/s^2 * t^2
t= 2sqrt(3) s?

4. May 11, 2014

### jackarms

It looks like you lost a t along the way -- when you have the 28 m/s, there's also a t multiplied behind it, if you look at the equation above it.

I'd also recommend to leave everything as a variable until the very end, since this usually makes it easier to solve, and you don't have to worry about writing down potentially large numbers at every step.

So if you start here:
$$v_{t}t = v_{c}t + \frac{1}{2}at^{2}$$
Where $v_{t}$ is the speed of the truck, and $v_{c}$ is the speed of the car, then isolate t (so that you have it in terms of all the other variables), then you can substitute.

And if you want to do the fancy writing for equations and writing, just put all of the text you want modified between [ itex ] and [ /itex ], but without the spaces.

So the raw text you type in looks like:
[ itex ] stuff [ /itex ]
But without the spaces around itex and /itex.

You can read more here -- https://www.physicsforums.com/showthread.php?t=8997 -- but don't worry about learning a whole new language just to ask a physics question.

5. May 11, 2014

### Sullivan

That looks like a language very well worth learning. Thanks for introducing it to me. I'll do some reading when I'm done this assignment :]

Thanks for spotting my error.

So I manipulated the v(truck) * t = v initial (car) * t + 1/2 a * t^2 and got 2 * (v (truck) - v initial (car)) / a = t

and plugged everything in at the end like you said and got t= 12s

*plug that back into the formula for displacement and get 480 m as displacement.

use final velocity^2 = initial velocity^2 + 2a∆x to get final velocity, which comes out to 59.0 m/s

I think that's it! ..?

Last edited: May 11, 2014
6. May 11, 2014

### jackarms

Yes, it's really worth it to learn, especially if you use the forums a lot. It just makes everything look nice.

The time and displacement are both right, but I got 52 m/s for the final velocity -- maybe you had a typo somewhere? But other than that, you're all done. Glad you were able to work it all out :)

7. May 11, 2014

### Sullivan

Ah. That's right. I think I input my fields wrong in my calculator. Yikes! I tried again and got 52 myself. Yay!

It's quite a good feeling when you get something right. Even though I recieved lots of help. Thanks again for helping me and the welcome to the forum!