# Thread: Simple Linear Motion Problem

1. ## Simple Linear Motion Problem

If I know the initial velocity and the distance traveled before coming to a stop (velocity = 0) and assuming the (negative) acceleration is constant, how can I calculate the acceleration?

E.G.
I'm in a buggy going 100 feet/sec and there is a light 33 feet in front of me. How much (constant negative) acceleration do I need to come to a stop after 33 feet?

I can solve other problems, but when I know only the initial velocity and the distance traveled I can't seem to figure out how to calculate the acceleration.

-Thanks!

2. Use

$v^2 = u^2 + 2 a s$,

where $v$ is final velocity (zero in this case), $u$ is initial velocity, $a$ is acceleration (negative here) and $s$ is the distance traveled.

3. Originally Posted by JamesLeighe
If I know the initial velocity and the distance traveled before coming to a stop (velocity = 0) and assuming the (negative) acceleration is constant, how can I calculate the acceleration?

E.G.
I'm in a buggy going 100 feet/sec and there is a light 33 feet in front of me. How much (constant negative) acceleration do I need to come to a stop after 33 feet?

I can solve other problems, but when I know only the initial velocity and the distance traveled I can't seem to figure out how to calculate the acceleration.

-Thanks!
$v^2=u^2+2as$

$0^2=(100)^2+2a(33)$

$a=\frac{-10000}{66}=-151.51\; ft/s^2$

4. Thanks guys!

Now I just have to look into how it's derived so I can understand why it works.

5. I guess you mean you want to know how the equation is derived.

For constant acceleration, $a$, we have that

$\frac{dv}{dt} = a$,

but using the chain rule we have that

$\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$

so

$v \frac{dv}{dx} =a$.

Integrating this wrt $x$ we have

$\int v \, \mathrm{d} v = a \int \, \mathrm{d}x$

and if we replace $v$ and $x$ with dummy variables $v'$ and $x'$ and use the fact that at $x'=0$, $v'=u$ and at $x'=s$, $v' = v$ to add the limits we have

$\int_{u}^{v} v' \, \mathrm{d} v' = a \int_{0}^{s} \, \mathrm{d}x'$

which gives us that

$\frac{1}{2} \left(v^2 - u^2 \right) = a s$,

$\Leftrightarrow v^2 = u^2 + 2 a s$.

Hope that helps!

6. Yeah, that's what I really needed. Thanks Doc.