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Finding acceleration at a given time

A particle moves along a straight line with position function s(t)=ln⁡(t2+1)s(t) = \ln(t^2 + 1), where tt is measured in seconds and ss in meters.

Find the acceleration of the particle at t=3t = 3 seconds.

Hint

First find v(t)=s′(t)v(t) = s'(t) using the chain rule, then differentiate again to get a(t)=v′(t)a(t) = v'(t). Use the quotient rule for the second derivative, then substitute t=3t = 3.

Step-by-step solution

Differentiate s(t)=ln⁡(t2+1)s(t) = \ln(t^2+1) using the chain rule: v(t)=s′(t)=2tt2+1v(t) = s'(t) = \dfrac{2t}{t^2+1}.

Differentiate v(t)v(t) using the quotient rule: a(t)=v′(t)=2(t2+1)−2t(2t)(t2+1)2=2−2t2(t2+1)2a(t) = v'(t) = \dfrac{2(t^2+1) - 2t(2t)}{(t^2+1)^2} = \dfrac{2 - 2t^2}{(t^2+1)^2}. Substituting t=3t = 3 gives a(3)=2−18100=−16100=−0.16a(3) = \dfrac{2 - 18}{100} = \dfrac{-16}{100} = -0.16 m/s2^2.

Concavity of the position function

Using your formula for a(t)=s′′(t)=2−2t2(t2+1)2a(t) = s''(t) = \dfrac{2 - 2t^2}{(t^2+1)^2}, determine whether the graph of the position function s(t)s(t) is concave up or concave down at t=3t = 3, and justify your answer using the sign of s′′(t)s''(t).

Hint

Concavity of s(t)s(t) is determined by the sign of s′′(t)=a(t)s''(t) = a(t): positive means concave up, negative means concave down. Evaluate the sign of a(3)a(3) found in part (a).

Step-by-step solution

The concavity of s(t)s(t) is governed by the sign of its second derivative, s′′(t)=a(t)s''(t) = a(t). If a(t)>0a(t) > 0 the graph is concave up, and if a(t)<0a(t) < 0 the graph is concave down.

From part (a), a(3)=−0.16<0a(3) = -0.16 < 0, so the graph of s(t)s(t) is concave down at t=3t = 3.

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