Distance traveled during changing acceleration (calculus-based physics)

Question

The acceleration as a function of time $a(t)$ (in m/s$^2$) and the initial velocity $v(0)$ are given for a particle moving along a line: $$a(t) = 2t + 4, \hspace{4mm}v(0) = −5, \hspace{8mm} 0 \leq t \leq 4.$$ (a) Find the velocity at time $t$. ($v(t) = t^2+4t−5$)

(b) Find the distance traveled during the given time interval.

I was able to solve part (a) but have been having issues with part (b). It's a webassign assignment and the "master it" section only gives me the first portion and it hasn't allowed me to see other problems so I could try and figure it out by example. Is anyone able to explain to me how to do this?

Answer 1

Hint: The initial velocity is $v(0)=-5$ m/s, whereas from part (a) the final velocity is $v(4) = +27$ m/s. So the particle evidently moves to the left, comes to rest at some time $T$, and then moves back to the right until time $t=4$. How far does it move to the left by time $T$, and how far does it move to the right in the time remaining?

Answer 2

You have an expression for the velocity.

To calculate distance as opposed to displacement, you need to know the time(s) when $v=0$, since the particle reverses the direction of motion at these times, and integrate the velocity separately over these time intervals, taking the absolute value where necessary.

Clearly $v=0\Rightarrow t=1$

Therefore the distance travelled is $$\left|\int_0^1(t^2+4t-5)dt\right|+\left|\int_1^4(t^2+4t-5)dt\right|$$