Question

find the velocity and the acceleration at time t = 5 s for the position function s(t) = \sqrt{t^{2}+5}.

v(5) = \frac{\sqrt{30}}{6} \frac{m}{s}

a(5) = 2 \frac{m}{s^{2}}

a(5) = \square \square

Explanation:

Step1: Find the velocity function

The velocity \(v(t)\) is the derivative of the position function \(s(t)\). Given \(s(t)=\sqrt{t^{2}+5}=(t^{2}+5)^{\frac{1}{2}}\).
Using the chain rule \((u^{n})^\prime = nu^{n - 1}u^\prime\), where \(u=t^{2}+5\), \(n=\frac{1}{2}\), and \(u^\prime = 2t\).
So \(v(t)=\frac{1}{2}(t^{2}+5)^{-\frac{1}{2}}\times2t=\frac{t}{\sqrt{t^{2}+5}}\).
When \(t = 5\), \(v(5)=\frac{5}{\sqrt{5^{2}+5}}=\frac{5}{\sqrt{30}}=\frac{\sqrt{30}}{6}\text{ m/s}\).

Step2: Find the acceleration function

The acceleration \(a(t)\) is the derivative of the velocity function \(v(t)\).
Using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = t\), \(u^\prime=1\), \(v=(t^{2}+5)^{\frac{1}{2}}\), and \(v^\prime=\frac{t}{\sqrt{t^{2}+5}}\) (from Step1).
\(a(t)=\frac{\sqrt{t^{2}+5}-t\times\frac{t}{\sqrt{t^{2}+5}}}{t^{2}+5}=\frac{t^{2}+5 - t^{2}}{(t^{2}+5)^{\frac{3}{2}}}=\frac{5}{(t^{2}+5)^{\frac{3}{2}}}\).
When \(t = 5\), \(a(5)=\frac{5}{(5^{2}+5)^{\frac{3}{2}}}=\frac{5}{(30)^{\frac{3}{2}}}=\frac{5}{30\sqrt{30}}=\frac{1}{6\sqrt{30}}=\frac{\sqrt{30}}{180}\text{ m/s}^2\approx\frac{1}{33}\text{ m/s}^2\). But if there is a calculation error in the problem - setup (assuming a different approach, if we use the power - rule and chain - rule for \(v(t)=\frac{t}{\sqrt{t^{2}+5}}\)):
Another way, \(v(t)=t(t^{2}+5)^{-\frac{1}{2}}\). Using the product rule \((uv)^\prime=u^\prime v+uv^\prime\), \(u = t\), \(u^\prime = 1\), \(v=(t^{2}+5)^{-\frac{1}{2}}\), \(v^\prime=-\frac{1}{2}(t^{2}+5)^{-\frac{3}{2}}\times2t\).
\(v^\prime(t)=(t^{2}+5)^{-\frac{1}{2}}+t\times(-\frac{1}{2}(t^{2}+5)^{-\frac{3}{2}}\times2t)=\frac{t^{2}+5 - t^{2}}{(t^{2}+5)^{\frac{3}{2}}}=\frac{5}{(t^{2}+5)^{\frac{3}{2}}}\). When \(t = 5\), \(a(5)=\frac{5}{(25 + 5)^{\frac{3}{2}}}=\frac{5}{30\sqrt{30}}=\frac{1}{6\sqrt{30}}=\frac{\sqrt{30}}{180}\approx\frac{1}{33}\text{ m/s}^2\). If we assume the problem has a typo and we use the formula \(a(t)\) from the second - derivative of \(s(t)\) in a more straightforward power - rule application:
If \(s(t)=(t^{2}+5)^{\frac{1}{2}}\), \(s^\prime(t)=\frac{t}{\sqrt{t^{2}+5}}\), \(s^{\prime\prime}(t)=\frac{\sqrt{t^{2}+5}-\frac{t^{2}}{\sqrt{t^{2}+5}}}{t^{2}+5}=\frac{5}{(t^{2}+5)^{\frac{3}{2}}}\). When \(t = 5\), \(a(5)=\frac{5}{(25 + 5)^{\frac{3}{2}}}=\frac{5}{30\sqrt{30}}=\frac{1}{6\sqrt{30}}=\frac{\sqrt{30}}{180}\approx\frac{1}{33}\text{ m/s}^2\). But if we consider the following:
If \(s(t)=\sqrt{t^{2}+5}\), \(v(t)=\frac{d s}{dt}=\frac{t}{\sqrt{t^{2}+5}}\), \(a(t)=\frac{d v}{dt}=\frac{\sqrt{t^{2}+5}-t\times\frac{t}{\sqrt{t^{2}+5}}}{t^{2}+5}=\frac{5}{(t^{2}+5)^{\frac{3}{2}}}\). Substitute \(t = 5\) into \(a(t)\):
\(a(5)=\frac{5}{(25 + 5)^{\frac{3}{2}}}=\frac{5}{30\sqrt{30}}=\frac{1}{6\sqrt{30}}=\frac{\sqrt{30}}{180}\approx\frac{1}{33}\text{ m/s}^2\). If we assume the problem has a miscalculation in the given \(v(5)\) (maybe rational - izing the denominator wrong), but following the correct calculus:
\(v(5)=\frac{5}{\sqrt{30}}=\frac{\sqrt{30}}{6}\text{ m/s}\), \(a(5)=\frac{5}{(25 + 5)^{\frac{3}{2}}}=\frac{5}{30\sqrt{30}}=\frac{1}{6\sqrt{30}}=\frac{\sqrt{30}}{180}\text{ m/s}^2\).

Answer:

\(v(5)=\frac{\sqrt{30}}{6}\text{ m/s}\), \(a(5)=\frac{\sqrt{30}}{180}\text{ m/s}^2\)