# Speed of a Braking Car

1. Sep 14, 2015

### r26h

1. The problem statement, all variables and given/known data
A car comes to a stop in 7.5 meters while braking at a rate of 2.0m/s each second.
A) What was the initial speed of the car?
B) What is the speed of the car after 0.5 seconds?
2. Relevant equations
Final velocity = Initial velocity + acceleration*time

3. The attempt at a solution
A) 0=Initial velocity+(-2)(7.5)
Initial velocty = 15 m/s

B)0=15+(-2)(.5)
Velocity=14m/s

Last edited by a moderator: Sep 14, 2015
2. Sep 14, 2015

### SteamKing

Staff Emeritus
Braking at a rate of 2.0 m/s each second sounds like a deceleration.
Did you check the units of this calculation? I don't think 2.0 m/s per second multiplied by 7.5 meters will give an answer in seconds
A "breaking" car is on which goes to pieces. A "braking" car is one which comes to a stop using its brakes.

3. Sep 14, 2015

### Staff: Mentor

This equation requires the stopping time; you are told the stopping distance.

4. Sep 14, 2015

### r26h

Ok, so what if to find the time I did 7.5m/ 2s/m = 3.75 seconds.

Then used that equation:
0=Initial velocity + (-2m/s)(3.75s)
So the initial speed is 7.5m/s? Does that make sense?

5. Sep 15, 2015

### haruspex

I guess you mean 7.5m/(2m/s), but you must not go plugging values into equations just because you know an equation that wants inputs of that type. You have to think about whether the equation is appropriate to the circumstance. What standard equation did you invoke to arrive at that calculation? What do the variables in the equation represent?

6. Sep 15, 2015

### r26h

Ok, I see my error. What if I used Vf^2=Vi^2+2ax, rearranged to Vi=Square root of (Vf-2ax)
=square root (0-2(-2)(7.5)
initial velocity=square root of 30 or about 5.48 meters/second

7. Sep 15, 2015

### r26h

To find the time, next I used Vf=Vi+at
0=5.48m/s +(-2m/s)(t)
-5.48m/s==2t
t=2.74s

Then which equation would I use to find the speed at 0.5 seconds?

8. Sep 15, 2015

### haruspex

You are not asked for the time to stop, are you? But the equation you used will also tell you the speed after a given time.