Question

A car moving at a constant rate decelerates and comes to a halt over a distance of 125m. If the car decelerates at 1.5m/s^2, how long does it take the car to stop?

I'm thinking I should use the equation: V final^2 = V intitial*t + 1/2at^2

But my acceleration is negative and I can't take the square root of a negative.

Answer 1

The trouble here is that if you want to use that equation (which by the way should be d = V intitial*t + 1/2at^2), you need to know the value of the initial velocity, which the question does not give you.

You need an equation that uses distance, acceleration, and final velocity (since those are given; note final velocity is 0 m/s) and time (which you'll have to solve for). We can come up with this equation using two equations you already know:

\[(1) \space a = \frac{v_f-v_i}{t} \implies v_i=v_f-at\]\[(2) \space v_f^2=v_i^2+2ad\]
Notice I rearranged equation (1) to solve for initial velocity - this way everything on the right side is stuff we're interested in. Now I substitute this value into equation (2):

\[v_f^2=(v_f-at)^2+2ad\]
Now you can plug in all your numbers and solve for t!
\[0=(0+1.5t)^2+(2)(-1.5)(125)\]\[2.25t^2=375\]\[t^2=166.67\]\[t \approx 13\]
So it takes about 13 seconds for the car to come to a full stop! let me know if you have any questions.