A particle moves along a curve so that its velocity is given by $v(t)= t^2 +1t −12$ A particle moves along a curve so that its velocity (in meters per second) is given by
$$v(t)= t^2 +1t −12$$
$(a)$ Find the displacement over the time interval $[0,4]$
$(b)$ Find the distance over the time interval $[0,4]$
$(c)$ Find the average speed over the time interval
$[0,4]$
$(d)$ Find the avrage velocity over the time interval
$[0,4]$
I don't know how to do the last three, but for $(a)$, I used the antiderivative, and found the answer to be $-56/3$.
Could someone help with the process for the last three?
 A: For b: The difference between distance and displacement is that when calculating distance, you account for moving in the negative direction, while displacement only takes the initial and final placement into account.
For example, if I walk forward 10 feet, then walk back 5 feet, my displacement is 5 feet, but my total distance is 15 feet.
This means we count every step as a positive velocity, or we take the absolute value of the velocity. Therefore: to calculate distance over a time interval [a,b], we take the integral of |v(t)|dt from a to b.
To take the integral of the absolute value of the function, first factor the function and find where the function is negative on the interval using the zeros. Then, you can split the integral at that zero, and multiply the interval where the function is negative by -1 to make it positive.
For c: Speed is the absolute value of velocity. Therefore, the average speed over [0,4] is 1/4 * the integral from 0 to 4 of |v(t)|dt. You can use your answer in part b to calculate this quickly.
For d: Same process as c, except without the absolute value.
