# Uniform Accelerated Motion

1. Feb 3, 2013

### Nirupt

1. The problem statement, all variables and given/known data

A car decelerates uniformly and comes to a stop after 10 s. The car's average velocity during deceleration was 50 km/h. What was the car's acceleration while slowing to a stop?

{Since the acceleration is UNIFORM (constant), use vavg = (v0 + v)/2..... }

2. Relevant equations

3. The attempt at a solution

V0 is not known, Final V = 0, t = 10, Vavg is 13.9 m/s and trying to find a

But when I try to plug it in I feel I have to solve for v0

Last edited: Feb 3, 2013
2. Feb 3, 2013

### tms

Start by writing down the equations that describe the car's motion.

3. Feb 3, 2013

### Nirupt

The equation I come up with was 13.9(10) = (v0 - 0/2)(10)
But that equals to -27.8 m/s^2 which is not the right answer

4. Feb 3, 2013

### SammyS

Staff Emeritus
Do it one step at a time.

You have, vavg = (v0 + v)/2 , and you know vavg ≈ 13.9 m/s and v = 0 . So solve that for v0 .

After that it's pretty straight forward to find the acceleration.

5. Feb 3, 2013

### tms

Write the equations using symbols, not numbers. It is easier to see what is happening with the physics that way.

6. Feb 3, 2013

### Nirupt

So 13.9m/s = (v0 + 0)/2 = 23.7 m/s
V0≈23.7 m/s
V≈0
t≈10
Then Δv/Δt..

0-23.7m/s
-----------
10 s

= -2.37 m/s2

7. Feb 3, 2013

### SammyS

Staff Emeritus
Looks good !

8. Feb 4, 2013

### hms.tech

No, there still seems to be an arithmetic mistake :tongue:

$V_{0}$$\approx$27.8 m/s

9. Feb 5, 2013

### Nirupt

I'm not sure why I typed that but I know the answer is correct.