A particle moves along a line with velocity v(t) = t^2 + t metres per second. Find its displacement during the time period from t = 1 to t = 2 seconds.

Step-by-step solution

1. 1

   Relate displacement to velocity. Displacement over $[a,b]$ is the definite integral $\int_a^b v(t)\,dt$; this is the net change in position, the inverse of the fact that $v=\frac{ds}{dt}$.

2. 2

   Check the sign of the velocity. On $[1,2]$ both $t^2$ and $t$ are positive, so $v(t)>0$ throughout. The particle never reverses, which means displacement and total distance travelled coincide here.

3. 3

   Write the integral. Displacement $=\displaystyle\int_1^2\left(t^2+t\right)dt$.

4. 4

   Antidifferentiate. $\int\left(t^2+t\right)dt=\frac{t^3}{3}+\frac{t^2}{2}$.

5. 5

   Evaluate at the endpoints. At $t=2$: $\frac{8}{3}+2=\frac{14}{3}$. At $t=1$: $\frac{1}{3}+\frac{1}{2}=\frac{5}{6}$. Subtracting, $\frac{28}{6}-\frac{5}{6}=\frac{23}{6}$.

6. 6

   Report with units. The displacement is $\frac{23}{6}\approx 3.833$ metres, which numerical quadrature of $t^2+t$ on $[1,2]$ reproduces as $3.8333$.