# Finding the velocity function

• Feb 12th 2009, 04:39 PM
fattydq
Finding the velocity function
I thought acceleration was the change in velocity or in other words the derivative of the velocity function, and thus am trying to solve the problem I've attached keeping this in mind but it's not working. Any helpful hints?
• Feb 12th 2009, 04:52 PM
skeeter
Quote:

Originally Posted by fattydq
I thought acceleration was the change in velocity or in other words the derivative of the velocity function, and thus am trying to solve the problem I've attached keeping this in mind but it's not working. Any helpful hints?

you forgot your constant of integration when you integrated to get v(t) ... determine it from the given initial condition.
• Feb 12th 2009, 04:53 PM
Nacho
A god way can be integrate, knowing:

$
v(t) = d'(t) \Rightarrow \int {v(t)dt} = d(t) + C
$
• Feb 12th 2009, 04:56 PM
Scott H
Two or more functions can have the same derivative. In our case, any function whose derivative is $t+8$ must be of the form

$v(t)=\frac{1}{2}t^2+8t+C.$

All we have to do now is figure out what $C$ must be. We are told that $v(0)=3$, so

$v(0)=\frac{1}{2}0^2+8\cdot0+C=C=3.$

Now that we know $C=3$, we can rewrite our formula for $v(t)$:

$v(t)=\frac{1}{2}t^2+8t+3.$

You were close, though. :)
• Feb 12th 2009, 04:56 PM
fattydq
ahh gotcha, but then how would I go the final part of the problem calculating the total distance?
• Feb 12th 2009, 05:05 PM
fattydq
I can't even figure out how total distance can be contrived from what's given...
• Feb 12th 2009, 05:28 PM
HallsofIvy
Just as acceleration is the derivative of velocity, so velocity is the derivative of distance. Just as velocity is the integral of acceleration, so distance is the integral of velocity.

Now that you know the velocity function is $\frac{1}{2}t^2+ 8t+ 3$ integrate that. Because you are asked for the distance covered during the time interval 0 and 10, d(0)= 0 to find C, the constant of integration and the find d(10). Equivalently evaluate d(10)- d(0) so the constant cancels. That last is the same as finding the definite integral $\int_0^{10} (\frac{1}{2}t^2+ 8t- 3)dt$
• Feb 12th 2009, 05:36 PM
fattydq
Thanks buddy
• Feb 12th 2009, 06:25 PM
skeeter
Quote:

Originally Posted by HallsofIvy
Just as acceleration is the derivative of velocity, so velocity is the derivative of distance. Just as velocity is the integral of acceleration, so distance is the integral of velocity.

to get the terminology straight ...

position (a vector) is the antiderivative of velocity (also a vector). distance is a scalar quantity.

a definite integral of velocity over an interval of time yields the displacement, or change in position over that interval of time.

distance traveled over an interval of time is the definite integral of speed (a scalar equal to the absolute value of velocity) over that time interval.