# A car traveling at 45 ft/sec decelerates at a constant 3 feet per second squared. How many feet does the car travel before coming to a complete stop?

### 1 Answer | Add Yours

Under constant acceleration, velocity as a function of time is given by

`v(t)=at+v_0,` where `v_0` is the initial velocity. Here we are given that `v_0=45` and `a=-3` (I won't clutter things up with units, every distance measurement is in feet and every time interval in seconds anyway). The car stops when `v=0,` so we solve

`0=-3t+45,` which means the car comes to a stop after 15 seconds. Now, the formula for distance traveled in a certain time under constant acceleration is given by

`d(t)=1/2 at^2+v_0t.` Here `a` and `v_0` are the same as before, and when `t=15`, the distance is

`1/2(-3)(15)^2+45(15)=337.5` feet.

This is the way someone would do it "from scratch", however there is a slightly shorter way. There is a formula that says that

`v_f^2=v_i^2+2ad` , where `v_f` and `v_i` are the final and initial velocities, respectively, `a` ` ` is acceleration, and `d` is distance. So for us, `v_f=0,` `v_i=45,` `a=-3` , and we can solve for `d` to get, as before,

**the distance traveled is 337.5 feet.**

**Sources:**