# Finding velocity as a function of position, constant acceleration

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!

PeroK
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The time it takes a truck to stop is v/a. The average speed during this time is v/2. So the distance travelled when the truck stops is $$s = v^2/2a$$
And, if the truck is still travelling 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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The time it takes a truck to stop is v/a. The average speed during this time is v/2. So the distance travelled when the truck stops is $$s = v^2/2a$$
And, if the truck is still travelling 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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PeroK
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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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PeroK
Homework Helper
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2020 Award
Are you trying to find speed as a function of position, or initial speed as a function of total stopping distance?

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.

PeroK
Homework Helper
Gold Member
2020 Award
You need all your quantities expressed in the same units: fps. Do that before you plug anything into your equation.

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.

PeroK