Particle on a line question with integrals calc2

Question

I have a question for calculus two regarding a particle on a line.

Question: A particle moves along a line with acceleration $a(t)=−1(t+3)~2\text{ft}/\text{sec}^2$. Find the distance traveled by the particle during the time interval $[0,1]$, given that the initial velocity $v(0)$ is $13~\text{ft/sec}$.

I think I'm supposed to use integrals, but I'm quite sure how to go about it. If anyone could help me out I'd be grateful, thank you <3

Answer 1

The velocity over $[0,1]$ is $$ v(t) = v(0) + \int_{0}^{t} a(x) \ dx $$ The distance (not displacement) over this interval is $$ \int_{0}^{1} |v(t)| \ dt $$ So you can find the distance with two integrals. It looks like your $a(t)$ is negative over $[0,1]$, but the initial velocity is much higher than $\int a(t) \ dt$ on that interval.

Answer 2

Let $v(t) = \int_0^ta(x)dx$, thus we have $v(t) = -\frac{1}{2}(t+3)^2+c$. Now we plug in $v(0) = 13$ and we get that: $v(t) = -\frac{1}{2}(t+3)^2+21$. If we repeat this progress for $s(t)$, we get: $s(t) = \int_0^tv(x)dx$ and notice that $s(0) = 0$, implies: $s(t) = -\frac{1}{6}(t+3)^3+21t$. And the answer would be: $s(1)=10\frac{1}{3}$