Simple Uniform Acceleration Question

[wwwwww]

Homework Statement

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?

Homework Equations

V_f=V_i+at
d=V_it+.5at

The Attempt at a Solution

Well I tried to use the equation d=(V_f+V_i)/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.

 

[PeterO]

Homework Statement

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?

Homework Equations

V_f=V_i+at
d=V_it+.5at

The Attempt at a Solution

Well I tried to use the equation d=(V_f+V_i)/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.

why not use the formula

d = 1/2(V_f+V_i).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.

 

[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.

 

[SammyS]

why not use the formula

d = 1/2(V_f+V_i).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.

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(V_f + V_i).​

 

[wwwwww]

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

 

[SammyS]

Correct.

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

 

[PeterO]

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

Your answer is correct. If that is the answer printed in the book it is wrong.
If they were trying to express to 3 significant figures it should have been 1.14 x 10³ m or perhaps 1.14 km

 

[PeterO]

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(V_f + V_i).​

I didn't just throw in that equation.
There are 5 equations covering uniformly accelerated motion. wwwwww had listed 2 of them, I brought in a 3rd, more directly applicable, one and there are 2 more to go, but neither of them has direct application to this question.

Besides when wwwwww had listed d = (V_f + V_i)/2 it looked like a mis-quoting of the equation of motion I referred to.