# Homework Help: Simple Uniform Acceleration Question

1. Sep 23, 2011

### wwwwww

1. The problem statement, all variables and given/known data
Hello new user here. I'm utterly confused about uniform acceleration as my book does a very poor job (in my opinion) of explaining it. Could somebody please help me with this problem?

A spaceship far from any star or planet experiences uniform acceleration from 65.0m/s to 162.0m/s in 10.0s. How far does it move?

2. Relevant equations
Vf=Vi+at
d=Vit+.5at

3. The attempt at a solution
Well I tried to use the equation d=(Vf+Vi)/2 but I checked my answer with my teacher's answers and the answers were way different. I think it's because the problem uses uniform acceleration and not constant acceleration.

2. Sep 23, 2011

### PeterO

why not use the formula

d = 1/2(Vf+Vi).t

You seem to have forgotten there is a time factor in the formula.

Of course you could easily use the first two you quoted. Use the first to find the acceleration, then the second to use that acceleration to find the distance.

Final point: Uniform acceleration and Constant acceleration are the same thing.

3. Sep 23, 2011

### Xyius

Uniform acceleration and constant acceleration are the same thing, it means that the acceleration remains at a fixed value throughout the course of motion.

The problem is giving you two velocities, since it is stated that the acceleration is constant, you must compute the acceleration from the two velocities.

So you need to ask yourself, what is acceleration? It is the change in velocity over a period of time. This should give you everything you need to answer the question. :]

EDIT: In physics, you shouldn't just try to throw a formula at things, but rather reason it through and make sense of the formulas.

4. Sep 23, 2011

### SammyS

Staff Emeritus
The formula that PeterO threw at this problem is simply another way to write
displacement = (average velocity) ✕ time .​
In the case of uniform acceleration, it's true that
(average velocity) = 1/2(Vf + Vi).​

5. Sep 23, 2011

### wwwwww

Oh thanks for the help. I got 1135 m, and the answer in the book is 1140 which I guess is close enough.

6. Sep 23, 2011

### SammyS

Staff Emeritus
Correct.

(Looks like they rounded off to 3 significant figures.)

7. Sep 24, 2011