A bird flies for 8 seconds with a velocity of $|t-2|$ m/s in a straight line. Find distance covered by the bird.

Question

I have a tiny discrepancy which I knew how to solve, but seem to have forgotten.

$$x=\int t-2$$

I know we have to separate sovle the integral from 0 to 2 and then 2 to 8.

But in the second integral, do I put the lower limit as 2 or 3?

I tried putting 2 as the lower limit for the second integral, which gave me $16$, but the right answer is $2$

The things that’s throwing me off is modulus. Because of that, the bird actually never travels in reverse, and after two seconds , the variation in velocity is linear

Answer 1

Assuming the variable $t$ denotes the time in seconds since the bird has started flying, the distance $d$ in meters traveled is given by $$d=\int_0^8|t-2|\,{\rm d}t.$$ The integrand $|t-2|$ is piecewise linear, and so indeed it makes sense to split the integral into two parts as you did: $$\int_0^8|t-2|\,{\rm d}t=\int_0^22-t\,{\rm d}t+\int_2^8t-2\,{\rm d}t.$$ From here it is an elementary exercise in integration; we get \begin{eqnarray*} \int_0^22-t\,{\rm d}t&=&\left[2t-\tfrac12t^2\right]_0^2 &=&(2\cdot2-\tfrac12\cdot2^2)-(2\cdot0-\tfrac12\cdot0^2)&=&2-0&=&2,\\ \int_2^8t-2\,{\rm d}t&=&\left[\tfrac12t^2-2t\right]_2^8 &=&(\tfrac12\cdot8^2-2\cdot8)-(\tfrac12\cdot2^2-2\cdot2)&=&16-(-2)&=&18, \end{eqnarray*} so all in all we find that $d=2+18=20$.

Answer 2

Recall : $s=\displaystyle {\int} vdt$, where we have :

$t[sec]$, $v[m/s]$, and $s[m].$

$v_1(t)=2-t$ for $0 \le t \le 2;$

$v_2(t)= t-2$ for $2 \le t \le 8;$

$s= \displaystyle{\int_{0}^{2}}v_1(t)dt +\int_{2}^{8}v_2(t)dt$.

Can you finish ?