How do I find the total distance travelled by this particle in Calculus?
I am currently working on this question and I am at the final step, but I can't seem to get it.
So I have the velocity formula
$$3t^2 - (15t/2) + 3$$
I am trying to find the total distance traveled in interval $$0 <= t <= 4$$
I have already set the velocity equation to zero and I used the quadratic formula and arrived at the answers 1/2 and 2 that make the equation 0.
From there I did |v(0) - v(1/2)| + |v(1/2) - v(2)| + |v(2) - v(4)|
Which I know is not right because I didn't get the answer which is according to my Calculus prof is:
155/8
I am weak in physics so I never really got the grasp of these things. Can someone help me understand how to solve this part?
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$\begingroup$The distance is the integral of the speed. The speed is the absolute value of the velocity. As the velocity is positive over $[0,1/2)$ and $(2,4]$, you are expected to do $\int_0^{1/2} (3t^2-15t/2+3) dt+\int_{1/2}^2 -(3t^2-15t/2+3) dt+\int_2^{4} (3t^2-15t/2+3) dt$
$\endgroup$ 3 $\begingroup$If $s(t)$ is distance traveled from $0$ to $t$ and $v(t)$ is a velocity for moment of time $t$, then relations between them may be described as follows: $$v(t) = \frac{ds(t)}{dt}$$ $$s(t) = \int_0^t v(T)dT$$
So if you have $v(t)$, you should integrate it by $t$ to get $s(t)$. Hope you are able to calculate the integral by yourself.
$\endgroup$ 1 $\begingroup$Velocity is measured in $\frac{\text{meters}}{\text{second}}$, so right there, that can be a strong indication that you are going to need to multiply by seconds in order to get the distance, so you have $$v(t) = 3t^2 - \frac{15}{2}t + 3,$$ and this is in meters per second. But this isn't a constant velocity. It's always changing from $t = 0$ to $t = 4$, so it won't suffice to simply multiply by $4$ seconds. You need the change in the position function, or in other words, "How much did the position function change in those 4 seconds?"
Fortunately, you know that The definition of velocity is $$v = \frac{dx}{dt}$$ where $x$ is the position function. This means that $$x = \int v \ dt$$ So you know have the position as a function of time, so now you can find the change in position:
$$x(t) = \int 3t^2 - \frac{15}{2}t + 3 \ dt = t^3 - \frac{15}{4}t^2 + 3t$$.
But this gives the displacement, not the distance. So you need to find the zero of the velocity function (in the interval), which is $t = 2$. So just break the integral up into three parts: $|x(4) - x(2)| + |x(2) - x(1/2)| + |x(1/2) - x(0)|$.
The problem with your answer is that you found the changes in the velocity function, but you need to look at the changes in the position function in order to find the distance traveled.
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