# Simple Kinematics Question: Equation of Motion?

1. Sep 11, 2010

Hi everyone. I'm new here, and I would like some help understanding the equation of motion. It's a simple high school homework question, so I think you guys should be able to help me out here. I'm not sure if I'll be able to properly fit the question into the template, because I know how to answer the actual question... I'm just confused as to how or why it works the way it does.

1. The problem statement, all variables and given/known data
I need to find the position from X=0.

2. Relevant equations
So what I did is simply plug in the equation of motion.

Xf=X0+V0t+1/2at^2

So that gives me:

Xf=5m+3m/s(11s)+1/2(2m/s^2)(3s^2)

3. The attempt at a solution

If I simply plug in the numbers, I get X=41m. But I'm pretty sure this is wrong. I think that it's fairly clear that I'm using the equation of motion wrong, because logically, it doesn't make sense.

So let's start by looking at the first part of the equation, Xf=X0+V0t.

Okay, that part makes sense. The final position is where you started plus how far you went.

But acceleration is where I don't quite catch on. First of all, shouldn't the amount of time spent accelerating be distributed to the velocity, since time spent accelerating is also time spent traveling at the original velocity? In other words, when the car accelerates to 2m/sec/sec, that time is also spent traveling at the original velocity, just with the acceleration included. So why isn't the original velocity added to the acceleration in the equation of motion?

And second, I don't understand the acceleration part of the equation itself. In this particular problem, the car accelerates at 2m/sec/sec for 3 seconds. So that means that it's speed should increase by 2, then 4, then 6, for a total of 12 extra meters. But the equation of motion's formula is 1/2(2m/s^2)(3s^2), which just gives me 3 meters.

So clearly, I either have something wrong, or I'm misunderstanding how the equation of motion is supposed to work. Can anyone shed some light on this problem?

Thanks.

2. Sep 11, 2010

### ehild

The first part of the motion is uniform. Determine how far the car reaches during the first 8 s. You can apply the equation for motion with constant acceleration in the last 3 s.

ehild

3. Sep 11, 2010

Thanks but I still don't get why the formula for acceleration is 1/2at^2? It doesn't match up with what the numbers should logically be.

4. Sep 11, 2010

### ehild

It is Xf=X0+V0t+1/2at^2, but X0 is the position of the car when it starts to accelerate and V0 is the velocity at the same time.

ehild

5. Sep 11, 2010

### regarun

Xf=X0+V0t+1/2at^2

See here the part V0t is the distance covered by the car due to its initial velocity before acceleration starts and 1/2at^2 is the distance covered while uniform acceleration.

For the above question x=5+33+9=47m and x0+47 = 5 + 47 = 52 m

6. Sep 12, 2010

But how can it start accelerating at X0 when it doesn't accelerate (in the word problem) until after it's finished 8 seconds of traveling at the base velocity?

Please excuse me if I'm making myself look like an idiot, keep in mind that I'm very new to the mathematical aspects of physics.

7. Sep 12, 2010

### ehild

This is not the same X0 as the position at the beginning of the motion.
Tell me please what is the position of the car after 8 s? x(8)=?

ehild

8. Sep 14, 2010