← Back to learning objective

Net displacement from velocity

A particle moves along a straight line with velocity v(t)=4−t2v(t) = 4 - t^2 (in meters per second) for tt in seconds, where t∈[0,3]t \in [0,3].

Find the net displacement of the particle over the interval [0,3][0,3], i.e., compute ∫03v(t) dt=∫03(4−t2) dt.\int_0^3 v(t)\,dt = \int_0^3 (4-t^2)\,dt.

Hint

Apply the Net Change Theorem: find an antiderivative F(t)F(t) of v(t)v(t), then evaluate F(3)−F(0)F(3) - F(0).

Step-by-step solution

An antiderivative of v(t)=4−t2v(t) = 4 - t^2 is F(t)=4t−t33F(t) = 4t - \dfrac{t^3}{3}. By the Net Change Theorem, ∫03v(t) dt=F(3)−F(0)\int_0^3 v(t)\,dt = F(3) - F(0).

Compute F(3)=12−9=3F(3) = 12 - 9 = 3 and F(0)=0F(0) = 0. So the net displacement is 3−0=33 - 0 = 3 meters.

Total distance traveled

Find the total distance traveled by the particle over [0,3][0,3]. Note that v(t)=4−t2v(t) = 4 - t^2 changes sign somewhere in this interval, so you will need to account for direction changes.

Hint

Find where v(t)=0v(t) = 0 on [0,3][0,3], split the integral there, and take the absolute value of each piece before adding them up.

Step-by-step solution

Set v(t)=4−t2=0v(t) = 4 - t^2 = 0, giving t=2t = 2 (the only root in [0,3][0,3]). On [0,2][0,2], v(t)≥0v(t) \geq 0, and on [2,3][2,3], v(t)≤0v(t) \leq 0, so the motion reverses direction at t=2t=2.

Using F(t)=4t−t33F(t) = 4t - \dfrac{t^3}{3}: on [0,2][0,2], displacement is F(2)−F(0)=(8−83)−0=163F(2)-F(0) = \left(8 - \frac{8}{3}\right) - 0 = \frac{16}{3}. On [2,3][2,3], displacement is F(3)−F(2)=3−163=−73F(3)-F(2) = 3 - \frac{16}{3} = -\frac{7}{3}.

Total distance is the sum of the absolute values of these two pieces: ∣163∣+∣−73∣=163+73=233\left|\frac{16}{3}\right| + \left|-\frac{7}{3}\right| = \frac{16}{3} + \frac{7}{3} = \frac{23}{3} meters.

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap