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Net displacement via the Net Change Theorem

A particle moves along a straight line with velocity function v(t)=e−t−e−2tv(t) = e^{-t} - e^{-2t} (in meters per second), for tt measured in seconds. The particle is decelerating over the interval [0,5][0,5].

Use the Net Change Theorem to find the net displacement of the particle over the time interval [0,5][0,5]. That is, evaluate

∫05(e−t−e−2t) dt\int_0^5 \left(e^{-t} - e^{-2t}\right)\, dt

and report the net displacement (include units).

Hint

Find an antiderivative of each term separately using ∫ekt dt=1kekt+C\int e^{kt}\,dt = \frac{1}{k}e^{kt} + C, then apply the Net Change Theorem by evaluating the antiderivative at t=5t=5 and t=0t=0 and subtracting.

Step-by-step solution

An antiderivative of v(t)=e−t−e−2tv(t) = e^{-t} - e^{-2t} is found term by term: ∫e−t dt=−e−t\int e^{-t}\,dt = -e^{-t} and ∫e−2t dt=−12e−2t\int e^{-2t}\,dt = -\tfrac12 e^{-2t}. So an antiderivative is s(t)=−e−t+12e−2t+Cs(t) = -e^{-t} + \tfrac12 e^{-2t} + C.

By the Net Change Theorem, the net displacement is s(5)−s(0)=(−e−5+12e−10)−(−1+12)=12−e−5+12e−10s(5)-s(0) = \left(-e^{-5}+\tfrac12 e^{-10}\right) - \left(-1+\tfrac12\right) = \tfrac12 - e^{-5} + \tfrac12 e^{-10}. Numerically this is approximately 0.49330.4933 meters.

Note the positive result confirms the particle ends up net displaced in the positive direction, even though its velocity is decreasing (decelerating) toward zero as tt increases.

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