# Simple velocity question

1. Jul 7, 2011

### rjs123

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

An object is traveling in a straight line with an initial velocity of 3 m/s and an acceleration a(t) = st, where s = 2 m/s ^3 and t is measured in seconds. Find a time T such that v(T) = 21 m/s.

I wanted to use v = vo + at ... to find t...but the function of a(t) is kind of confusing...also i just started physics a couple weeks ago...so bear with me.

2. Jul 7, 2011

### SammyS

Staff Emeritus
v = vo + at is true is acceleration is constant.

In general a(t), a as a function of t is given by, $\displaystyle a(t)=\frac{d}{dt} v(t)$, acceleration is the time derivative of the velocity.

Therefore, velocity is the anti-derivative of the acceleration.

$$v(t)=\int a(t)\,dt$$
You will need to evaluate the constant of integration by considering the initial conditions.

If you're more comfortable treating it as a definite integral, then $\displaystyle v(t)=\int_{t_0}^{t} a(t)\,dt \,.$

3. Jul 7, 2011

### rjs123

ok...where does the initial velocity get plugged into that equation though.

When you integrate dont you just get 1/2*at^2

I got caught up on the cubed part of the acceleration...because im used to seeing it squared...like 9.8 m/s ^2.

4. Jul 7, 2011

### tiny-tim

hi rjs123!

(try using the X2 icon just above the Reply box )
no

(if a is constant, and you integrate a twice, you get 1/2 at2)

you have dv/dt = st, so integrate to get v

5. Jul 7, 2011

### rjs123

i understand it goes

x(t) = position

x'(t) = velocity

x''(t) = acceleration

to get the velocity from the acceleration just integrate...i understand that...but setting up the problem is the thing i'm having difficulty with even though it is something simple...the thing that makes me troubled is that the acceleration isn't constant.

6. Jul 7, 2011

### Staff: Mentor

That's why you want to integrate the acceleration; because it is not constant but a function of time.

You have an expression for a(t), and you want to integrate a(t). So....

7. Jul 7, 2011

### Superstring

At time t=0, the velocity is 3 m/s. When you integrate the acceleration you get a new function of t plus a constant. What do you thing that the constant should be to satisfy the conditions?

If a didn't change with time (it was constant), then when you integrate you would get $\frac{1}{2}at^2$, but your acceleration DOES change with time so you will end up with something different.

The 's' part isn't the acceleration - 'st' together is the acceleration. When you multiply m/s3 by seconds, you get m/s2 which are the correct units for acceleration.

I think the 's' is what's throwing you off. Try forgetting about units for the time being - the acceleration function is really just a(t)=2t. That should look more familiar for integration.

8. Jul 7, 2011

### rjs123

ok so...if you integrate a(t) = 2t...just becomes $$v(t) = t^2$$...

v = vo + at

would this be the correct approach?

$$21 m/s = 3 m/s + 2t(t)$$
$$18 m/s = 2t^2$$
$$9 m/s = t^2$$
$$3 seconds = t$$

Last edited: Jul 7, 2011
9. Jul 7, 2011

### Superstring

You forgot about the constant of integration (and units). It should read:

$$v(t) = \frac{1}{2} st^2+C$$

Now you need to figure out what C is equal to with the initial condition that v(0)=3 m/s

You can't use that equation unless a is constant, which it is not. You ALREADY have an equation that tells you the velocity at any point in time (or, at least you will once you solve for C) - you just need to solve for t when v(t)=21.

10. Jul 7, 2011

### SammyS

Staff Emeritus
I see there have been quite a few posts since my first reply. But you seem to have missed the whole point of what I stated in the quoted text above.

You have in the original problem statement that the acceleration is given by: a(t) = s·t .

Therefore, taking the integral (anti-derivative) gives: $\displaystyle v(t)=\int st\,dt$, where s is a constant, namely s = 2 m/s3.

Therefore, v(t) = (1/2)s t2 + C .

At t=0, we have v(0) = 3 m/s. So, what is C ?

Using that value for C and plugging-in s = 2, should give you v(t) .

Then solve v(T) = 21 m/s for T .

11. Jul 8, 2011

### rjs123

thanks i got it now...I just started 8 days ago...so this is all new to me.

c = 3

21 = t^2 + c
18 = t^2
t = 4.2 s

12. Jul 8, 2011

Correct