Question

find the velocity and the acceleration at time $t = 5s$ for the position function $s(t)=sqrt{t^{2}+5}$. $a(5)=2\frac{m}{s^{2}}$ $v(5)=6\frac{m}{s}$ $v(5)=\square$

Explanation:

Step1: Find the velocity function \(v(t)\)

The velocity function \(v(t)\) is the derivative of the position function \(s(t)\). Given \(s(t)=\sqrt{t^{2}+5}=(t^{2}+5)^{\frac{1}{2}}\), use the chain rule \((u^{n})^\prime = nu^{n - 1}u^\prime\). Let \(u=t^{2}+5\), \(n=\frac{1}{2}\). Then \(u^\prime = 2t\). So \(v(t)=s^\prime(t)=\frac{1}{2}(t^{2}+5)^{-\frac{1}{2}}\times2t=\frac{t}{\sqrt{t^{2}+5}}\).

Step2: Calculate \(v(5)\)

Substitute \(t = 5\) into \(v(t)\). \(v(5)=\frac{5}{\sqrt{5^{2}+5}}=\frac{5}{\sqrt{25 + 5}}=\frac{5}{\sqrt{30}}=\frac{5\sqrt{30}}{30}=\frac{\sqrt{30}}{6}\approx1.29\) (This part seems there might be a mis - understanding. If we consider the formula \(v(t)=\frac{t}{\sqrt{t^{2}+5}}\), when \(t = 5\), \(v(5)=\frac{5}{\sqrt{25 + 5}}=\frac{5}{\sqrt{30}}=\frac{\sqrt{30}}{6}\). But if we assume the original position function is \(s(t)=\sqrt{t^{2}+5}\) and use the power - chain rule correctly \(v(t)=\frac{t}{\sqrt{t^{2}+5}}\), \(v(5)=\frac{5}{\sqrt{25+5}}=\frac{5}{\sqrt{30}}=\frac{\sqrt{30}}{6}\approx1.29\). However, if there is a typo in the problem and the position function is \(s(t)=\sqrt{t^{2}+5t}\), \(v(t)=\frac{2t + 5}{2\sqrt{t^{2}+5t}}\), \(v(5)=\frac{2\times5+5}{2\sqrt{25 + 25}}=\frac{15}{2\sqrt{50}}=\frac{15}{10\sqrt{2}}=\frac{3\sqrt{2}}{4}\approx1.06\). But following the given \(s(t)=\sqrt{t^{2}+5}\):

$$v(5)=\frac{5}{\sqrt{25 + 5}}=\frac{5}{\sqrt{30}}=\frac{\sqrt{30}}{6}$$

Answer:

\(\frac{\sqrt{30}}{6}\text{ m/s}\)