← Back to learning objective

Net displacement via integration of velocity

A particle moves along a straight line with velocity v(t)=5sin⁡(πt6)v(t) = 5\sin\left(\dfrac{\pi t}{6}\right) cm/s, where tt is measured in seconds.

Use the Net Change Theorem to find the net displacement of the particle over the time interval [0,12][0, 12]. That is, compute ∫0125sin⁡(πt6) dt.\int_0^{12} 5\sin\left(\frac{\pi t}{6}\right)\,dt.

Hint

Use u-substitution with u=πt6u = \dfrac{\pi t}{6}, so du=π6dtdu = \dfrac{\pi}{6}dt. Find the antiderivative first, then evaluate at the bounds t=0t=0 and t=12t=12 (or convert the bounds to uu-values and evaluate there instead).

Step-by-step solution

Let u=πt6u = \dfrac{\pi t}{6}, so du=π6 dtdu = \dfrac{\pi}{6}\,dt, which means dt=6πdudt = \dfrac{6}{\pi}du. Substituting gives ∫5sin⁡(u)⋅6π du=−30πcos⁡(u)+C=−30πcos⁡(πt6)+C\int 5\sin(u)\cdot\dfrac{6}{\pi}\,du = -\dfrac{30}{\pi}\cos(u) + C = -\dfrac{30}{\pi}\cos\left(\dfrac{\pi t}{6}\right) + C.

This antiderivative is F(t)=−30πcos⁡(πt6)F(t) = -\dfrac{30}{\pi}\cos\left(\dfrac{\pi t}{6}\right). By the Net Change Theorem, the net displacement is F(12)−F(0)F(12) - F(0). Since cos⁡(2π)=1\cos(2\pi) = 1 and cos⁡(0)=1\cos(0) = 1, we get F(12)−F(0)=−30π(1)−(−30π(1))=0F(12) - F(0) = -\dfrac{30}{\pi}(1) - \left(-\dfrac{30}{\pi}(1)\right) = 0.

Net displacement via integration of velocity — solution graph

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap