# Finding velocity as a function of position, constant acceleration

• crowned
In summary, a truck is traveling down the road at 60 mph, and begins to decelerate at a rate of 0.8 g. The truck slows to a stop in 150 feet using the equation v2 = 2a(s2-s1)+v12.
crowned
This isn't actually a homework question, it's for a vehicle safety project at work, but I think it fits in this forum.

## Homework Statement

A truck is traveling down the road at 60 mph, and begins to decelerate at a rate of 0.8 g.

I'd like to find the total distance it takes the truck to stop, and come up with a formula for the velocity as a function of displacement from the original braking point.

## Homework Equations

I think that the relevant equation is v2 = 2a(s2-s1)+v12

## The Attempt at a Solution

Well, I know the truck is traveling at 88 fps, and slowing at a rate of 0.8 * 32fps2, so it will take

88 fps / (32.2 fps2 * 0.8 g) * 44 fps = 150 feet to stop

But when I plug those numbers into the above equation, I get

v2 = (2 * (0.8 * 32.2 * 60 / 88) * (-x) + (60)2)0.5

That has the right y-intercept (65 mph), but the wrong x-intercept, around 104'

If I change the "2" on the right side of the equation to 1.38, it looks right, but I can't work out why it would be 1.38 instead of two.

Argh! Any help much appreciated!

The time it takes a truck to stop is v/a. The average speed during this time is v/2. So the distance traveled when the truck stops is $$s = v^2/2a$$
And, if the truck is still traveling at speed v1, then we have:
$$s_1 = (v^2-v_1^2)/2a$$
Where a is the magnitude of the deceleration and v is the initial speed.

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PeroK said:
The time it takes a truck to stop is v/a. The average speed during this time is v/2. So the distance traveled when the truck stops is $$s = v^2/2a$$
And, if the truck is still traveling at speed v1, then we have:
$$s_1 = (v^2-v_1^2)/2a$$
Where a is the magnitude of the deceleration and v is the initial speed.

Thanks for your response. However, I understand that part already--what I'm trying to work out is the equation for velocity as a function of position.

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You mean you understood all that but yet you couldn't rearrange the equation for yourself:

$$s_1 = (v^2-v_1^2)/2a$$
$$v_1^2 = v^2 - 2as_1$$
$$v_1 = \sqrt{v^2 - 2as_1}$$

Hi--thanks for the response. Well, I worked my way to the final equation you posted (it matches up to the final equation in my original post--may not have been clear, I didn't realize I could write in LaTeX). The problem is that I calculate a stopping distance of 150' by using average speed * time, but a "displacement when velocity equals zero" of 120' when I use the last of the equations you just posted.

Maybe I'm doing something else wrong? Here's the equation I'm using, with variables.

$$v_1 = \sqrt{(65mph)^2 - 2(0.8 g \cdot 32.2fps^2 \cdot 60mph/80fps) \cdot x}$$

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Are you trying to find speed as a function of position, or initial speed as a function of total stopping distance?

PeroK said:
Are you trying to find speed as a function of position, or initial speed as a function of total stopping distance?

Speed as a function of position. I'd like to know how fast the truck is going at a given displacement.

You need all your quantities expressed in the same units: fps. Do that before you plug anything into your equation.

PeroK said:
You need all your quantities expressed in the same units: fps. Do that before you plug anything into your equation.

Well, everything is expressed in mph.

Oh my gosh, how did that work!

Expressed in fps, it works fine.

y-intercept 88 fps, x intercept 150'

So to change the y-axis to mph, I just multiply the right side of the equation by 60mph/88fps, right?

Sheesh, didn't see that one coming.

1 mph = 5280/3600 fps = 88/60 fps, so yes.

Thanks so much for your help on this. Graphs are coming out perfectly.

## 1. What is the equation for finding velocity as a function of position with constant acceleration?

The equation for finding velocity as a function of position with constant acceleration is v = u + at, where v is the final velocity, u is the initial velocity, a is the constant acceleration, and t is the time.

## 2. How do you determine the direction of the velocity when using this equation?

The direction of the velocity can be determined by the sign of the acceleration. If the acceleration is positive, the velocity will be in the same direction as the initial velocity. If the acceleration is negative, the velocity will be in the opposite direction to the initial velocity.

## 3. Can this equation be used for both linear and circular motion?

Yes, this equation can be used for both linear and circular motion as long as the acceleration remains constant. For circular motion, the direction of the velocity may change, but the magnitude will remain constant.

## 4. How does the initial velocity affect the final velocity when using this equation?

The initial velocity has a direct impact on the final velocity. If the initial velocity is greater than the final velocity, the object will decelerate and come to a stop. If the initial velocity is less than the final velocity, the object will accelerate and increase in speed.

## 5. Is this equation only applicable to objects moving in a straight line?

No, this equation can also be used for objects moving in a curved path, as long as the acceleration remains constant. In this case, the velocity at any given point will be the tangent to the curved path at that point.

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