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?
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?
Two or more functions can have the same derivative. In our case, any function whose derivative is $\displaystyle t+8$ must be of the form
$\displaystyle v(t)=\frac{1}{2}t^2+8t+C.$
All we have to do now is figure out what $\displaystyle C$ must be. We are told that $\displaystyle v(0)=3$, so
$\displaystyle v(0)=\frac{1}{2}0^2+8\cdot0+C=C=3.$
Now that we know $\displaystyle C=3$, we can rewrite our formula for $\displaystyle v(t)$:
$\displaystyle v(t)=\frac{1}{2}t^2+8t+3.$
You were close, though.
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 $\displaystyle \frac{1}{2}t^2+ 8t+ 3$ 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 $\displaystyle \int_0^{10} (\frac{1}{2}t^2+ 8t- 3)dt$
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.