# What is the final velocity?

1. Sep 28, 2011

### albodibran

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

a car is at a velocity of 20 km/h if the car traveled 120 km in 3 hours at constant accelration. what is the final velocity?

2. Relevant equations

3. The attempt at a solution

I am completely confused. The book does not even list the particular formula above. And not much practice examples are given . I have an issue with understanding these variables. I dont see the correlation between the variables and the particular numbers. and its making a potentially easy problem non understandable.

2. Sep 28, 2011

### cepheid

Staff Emeritus
Re: Kinematics-velocity&acceleration

Welcome to PF albodibran!

In the equation you typed above, v is the current velocity (when the object is at distance d from its starting point), v0 is the initial velocity (when the object began accelerating). The variable 'a' is the acceleration (which is constant in this case) and d is the total displacement.

However, you don't need this equation. In fact, you don't even need the distance value you were given. Answer this question: what is the definition of acceleration?

Bearing in mind the definition above, if you have an acceleration value, and a time interval, what can those two things tell you about the velocity over that time interval?

3. Sep 28, 2011

### Ignea_unda

Re: Kinematics-velocity&acceleration

First, welcome to PF!

Let's see if we can help get you straightened out.

If we have constant acceleration,

$V= V_0 + at (1)$

$V$ is the final velocity, $V_0$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.

Also, we have the displacement,
$d = \frac{1}{2}(V + V_0) t (2)$

$d$ here, as said, is the distance traveled, also known as displacement.

If we rearrange (1) to solve for t, we get
$t = \frac{V-V_0}{a} (3)$

Now from here, just substitute (3) into (2) for t. You then have:
$d = \frac{1}{2}(V + V_0) (\frac{V-V_0}{a})$

When you work out your multiplication (I hope you're verifying what I'm doing here, it'll help) and rearrange things,

$V^2 = V_0^2 + 2ad$

Does that help you to see where they are coming from? If not, let us know and we'll see what we can do for you.

4. Sep 28, 2011

### albodibran

Re: Kinematics-velocity&acceleration

I figured that out thank you I will post a issue I have with part of another problem