A car has acceleration a(t) = t^2 - 2t + 3 metres per second squared at time t. If its velocity at t = 1 is 8 metres per second, find its velocity at t = 3.

Step-by-step solution

1. 1

   Use the relation between acceleration and velocity. Acceleration is the derivative of velocity, $a(t)=v'(t)$, so by the Fundamental Theorem of Calculus the change in velocity over an interval is the definite integral of $a$:

   $v(3)-v(1)=\int_1^3 a(t)\,dt$

   Working with the change (rather than finding $v$ with an unknown constant) means the initial condition is used cleanly at the end.

2. 2

   Antidifferentiate the polynomial term by term.

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

   No constant is needed because this antiderivative will be evaluated between limits.

3. 3

   Evaluate at the upper limit $t=3$.

   $\frac{27}{3}-9+9=9$

4. 4

   Evaluate at the lower limit $t=1$.

   $\frac{1}{3}-1+3=\frac{1}{3}+2=\frac{7}{3}$

5. 5

   Subtract to get the change in velocity.

   $\int_1^3 a(t)\,dt=9-\frac{7}{3}=\frac{27-7}{3}=\frac{20}{3}\ \text{m/s}$

6. 6

   Add the initial velocity.

   $v(3)=v(1)+\frac{20}{3}=8+\frac{20}{3}=\frac{24+20}{3}=\frac{44}{3}\approx 14.67\ \text{m/s}$

   A quick plausibility check: $a(t)$ stays between $2$ and $6$ on $[1,3]$, so the gain must lie between $4$ and $12$ m/s — and $\tfrac{20}{3}\approx 6.67$ does.