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Velocity and acceleration functions

A particle moves along a straight line with position function s(t)=1t+1−1s(t) = \dfrac{1}{t+1} - 1 for t≥0t \geq 0, where tt is measured in seconds and ss in meters.

Find the velocity v(t)v(t) and acceleration a(t)a(t) of the particle.

Hint

Rewrite s(t)=(t+1)−1−1s(t) = (t+1)^{-1} - 1 and use the power rule/chain rule to differentiate once for velocity and again for acceleration.

Step-by-step solution

Write s(t)=(t+1)−1−1s(t) = (t+1)^{-1} - 1. Differentiating using the chain rule (with inner function t+1t+1, whose derivative is 1), we get v(t)=s′(t)=−(t+1)−2=−1(t+1)2v(t) = s'(t) = -(t+1)^{-2} = -\dfrac{1}{(t+1)^2}.

Differentiating v(t)=−(t+1)−2v(t) = -(t+1)^{-2} again gives a(t)=v′(t)=2(t+1)−3=2(t+1)3a(t) = v'(t) = 2(t+1)^{-3} = \dfrac{2}{(t+1)^3}.

Long-term behavior of velocity and acceleration

Evaluate lim⁡t→∞v(t)\lim_{t \to \infty} v(t) and lim⁡t→∞a(t)\lim_{t \to \infty} a(t).

Hint

As t→∞t \to \infty, the denominators (t+1)2(t+1)^2 and (t+1)3(t+1)^3 grow without bound — think about what happens to a fraction with a fixed numerator and an ever-growing denominator.

Step-by-step solution

Since (t+1)2→∞(t+1)^2 \to \infty as t→∞t \to \infty, the fraction −1(t+1)2→0-\dfrac{1}{(t+1)^2} \to 0, so lim⁡t→∞v(t)=0\lim_{t\to\infty} v(t) = 0.

Similarly, (t+1)3→∞(t+1)^3 \to \infty, so 2(t+1)3→0\dfrac{2}{(t+1)^3} \to 0, giving lim⁡t→∞a(t)=0\lim_{t\to\infty} a(t) = 0.

Describing the long-term motion

Based on your results, describe in words what happens to the particle's motion as t→∞t \to \infty: consider the sign of v(t)v(t), the direction of motion, its position, and how its speed changes over time.

Hint

Check the sign of v(t)v(t) for all t≥0t \ge 0, note where s(t)s(t) tends as t→∞t\to\infty, and relate the decreasing magnitude of v(t)v(t) to the particle slowing down.

Step-by-step solution

Since v(t)=−1/(t+1)2<0v(t) = -1/(t+1)^2 < 0 for all t≥0t \geq 0, the particle always moves in the negative direction; it never stops or reverses.

As t→∞t \to \infty, s(t)=1/(t+1)−1→−1s(t) = 1/(t+1) - 1 \to -1, so the particle approaches the position s=−1s = -1 but never reaches it. Since ∣v(t)∣→0|v(t)| \to 0, the particle's speed continuously decreases toward 0, meaning it moves ever more slowly as it approaches this limiting position — essentially coming to rest asymptotically at s=−1s=-1.

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