# Formula given acceleration and distance

1. Dec 18, 2007

### Pure_Anarchy

1. The problem statement, all variables and given/known data
An object is pushed from rest across a sheet of ice, accelerating at 5.0 m/s over a distance of 80 cm. The object then slides at a constant speed for 4.0 s until it reaches a rough section that causes the object to stop in 2.5 s. (Assume 2 significant digits.)
(a) What is the speed of the object when it reaches the rough section?
(b) At what rate does it slow down once it reaches the rough section?
(c) What is the total distance that the object slides?

Can anyone tell me the formula I need to solve this question? Thanks in advance.

2. Relevant equations

3. The attempt at a solution

2. Dec 18, 2007

### Shooting Star

F = ma, where F is the TOTAL force acting on the body of mass m, as a result of which the body is having an accn of 'a'.

The three formulae for uniform accn, like vf = vi + at etc.

Last edited: Dec 18, 2007
3. Dec 18, 2007

### Jacque77

Hi Pure Anarchy. I'm a phy I student also. I took a shot at your problem. Did you get the same answer?
5m/s = initial acceleration.
0.8m = 80cm unit conversion

Vf = Vi +at
Vf = 0 (rest) + 5m/s * 0.8m = 4m/s
4m/s for 4 seconds = 16meters

Vf in the rough will be 0
0= 4m/s + a*2.5sec
-4 = a * 2.5
-4/2.5 = a
-1.6m/s^2 = a deceleration when it hits the rough

4m to stop is -1.6m/s^2*2.5s

16M+4M = 20Meters

4. Dec 19, 2007

### Shooting Star

>Vf = 0 (rest) + 5m/s * 0.8m = 4m/s

Do you think you've plugged in the correct value for t?