# Find constant acceleration with only initial speed and distance [closed]

Given the problem:

"A car moving initially at 50 mi/h begins decelerating at a constant rate 60 ft short of a stoplight. If the car comes to a stop right at the light, what is the magnitude of its acceleration?"

While this problem seems simple, I can't seem to find the correct formula to use. Most formulas I am finding require the use of time (t) which is not given in the problem statement. What formula(s) do I use to solve this problem? Am I supposed to use distance as the unit of time somehow? Or should I use some sort of derivation to get the number needed?

## closed as off-topic by John Rennie, ACuriousMind♦, user10851, Qmechanic♦Sep 6 '15 at 11:47

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• For constant acceleration you probably have a velocity vs. time relationship (formula, as you say), and the position vs. time. Use those two to eliminate the time variable and you will have velocity vs. position for a constant acceleration interval. It will really help in your studies to derive that formula yourself (although is can be found in a book, too ---but that doesn't help you learn it). – Bill N Sep 3 '15 at 22:11

Well let's pick the $(Ox)$ axis as a reference of frame and take the origin of time the instant the car starts deccelerating with a magnitude $a$:

Since we are talking about decceleration it is clear that:

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

So:

$$v(t)=v_0-at$$

And:

$$\frac{dx}{dt}=v_0-at\\ x(t)=v_0 t -\frac{a}{2}t^2$$

Now we know that the car will stop at the red light at some instant $t_a$, so:

$$v(t_a)=0\\ v_0-at_a=0\\ t_a=\frac{v_0}{a}$$

At the same instant $t_a$ the car would have rolled $d=60 ft$ as you stated, so:

$$x(t_a)=d\\ v_0t_a-\frac{a}{2} {t_a}^{2}=d\\ v_0\frac{v_0}{a}-\frac{a}{2}{\frac{v_0}{a}}^{2}=d\\ \frac{{v_0}^{2}}{a}-\frac{{v_0}^{2}}{2a}=d\\ \frac{{v_0}^{2}}{2a}=d\\ a=\frac{{v_0}^{2}}{2d}$$

Where $v_0=50mi/h$. Plug in the values and pay attention to the units in order to get your answer.

Note:

More generally we prove that for a uniformly deccelerating particle with acceleration $a$ that has initial velocity $v_i$ and final velocity $v_f$ that traveled a distance $d$ we obtain the following relation:

$$2ad={v_0}^{2}-{v_f}^{2}$$

There are four kinematic equations that apply to this type of problem. One of those equations can be used to solve for acceleration when you don't how long it took for the car to stop. The equation is:

$v_f^2 = v_i^2 + 2a \Delta\ x$

The initial velocity is given, the final velocity is zero, and the distance traveled is given. The only unknown, acceleration, is trivially easy to separate algebraically, and leads to an immediate answer for this type of problem.

The total distance traveled during the constant acceleration is 60 feet. Over the course of 60 feet the velocity has to go from 50mph to 0mph. That means the average velocity is 25mph. So, what amount of time going 25mph is equal to 60 feet? Then take that time and divide the change in velocity by it to arrive at the acceleration.