Mathematics

Mathematics

Problem

The velocity (in m/s) of a body is given by v(t)=ke−0.2t+3v(t)=ke^{-0.2t}+3 for 0≤t≤100\le t\le 10, where kk is a constant. Which expression equals the body's displacement (in meters) on this interval? a) 5k(1−e−2)5k\left(1-e^{-2}\right) b) 5k(e2−1)+305k\left(e^{2}-1\right)+30 c) 5k(1−e−2)+305k\left(1-e^{-2}\right)+30 d) 5k(e−2−1)+305k\left(e^{-2}-1\right)+30
Intuitive explanation

Displacement is the total distance traveled, found by integrating velocity over time—think of it as summing up all the tiny movements v(t) dtv(t) \, dt across the interval. Here you'll integrate v(t)=ke−0.2t+3v(t) = ke^{-0.2t} + 3 from t=0t=0 to t=10t=10. The key is handling the exponential term: remember that ∫eat dt=1aeat+C\int e^{at} \, dt = \frac{1}{a}e^{at} + C, so the constant kk and the coefficient −0.2-0.2 will both affect how the exponential integrates.

Intuitive explanation

Displacement is the total distance traveled, found by integrating velocity over time—think of it as summing up all the tiny movements v(t) dtv(t) \, dt across the interval. Here you'll integrate v(t)=ke−0.2t+3v(t) = ke^{-0.2t} + 3 from t=0t=0 to t=10t=10. The key is handling the exponential term: remember that ∫eat dt=1aeat+C\int e^{at} \, dt = \frac{1}{a}e^{at} + C, so the constant kk and the coefficient −0.2-0.2 will both affect how the exponential integrates.

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