# Finding velocity/distance from acceleration

1. May 9, 2012

### Tankertert

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

Hi guys. New here.

Im doing some last minute revision and i have a question im stuck on.

an objects acceleration a in ms^-1 at time t seconds is shown by

a(t) = 6t + ∏cos(2∏ t), where t ≥ 0

if objects initial velocity is 3ms^-1, find:

a) objects velocity after 2 seconds

b) distance object travels in first 5 seconds.

If you can show me your answers and how you solved the question so i can see how you did it, that would be amazing. thank you!

2. Relevant equations

3. The attempt at a solution

i firstly tried substituting the time in to solve, but that only gives me the acceleration and i need the velocity. and i have no idea on how to try and even get the second one.

2. May 9, 2012

### Infinitum

Hi Tankertert, welcome to PF!

It would be better if you show us what you did, and we try guiding you to the answer.

Do you know basic calculus? If yes, what is the definition of instantaneous acceleration? and instantaneous velocity?

3. May 9, 2012

### Tankertert

All i have done was sub 2 into a(t) = 6t + ∏cos(2∏ t) but i dont know how that would help me.

acceleration is velocity over time yes?
and velocity is distance over time?

4. May 9, 2012

### Matt4936

For the first question you would multiply the acceleration by the time.

For the second question you would get the velocity and multiply that by the time.

Also acceleration is ms^-2 not -1. I'm sure that was just a typo but I'm just making sure.

5. May 9, 2012

### Matt4936

For the first question you would multiply the acceleration by the time.

For the second question you would get the velocity and multiply that by the time.

Also acceleration is ms^-2 not -1. I'm sure that was just a typo but I'm just making sure.

6. May 9, 2012

### Infinitum

Although true, those are a bit vague definitions. Can you express the same mathematically, for instantaneous velocity/accelerations?

7. May 9, 2012

### Infinitum

This would work if the acceleration was constant, which it isn't....

8. May 9, 2012

### Matt4936

That means that you would have to integrate the acceleration function to get the velocity function and then substitute time in?

9. May 9, 2012

There we go!