Chapter 2, Problem 1CRE

To determine

Find the average velocity and instantaneous velocity of the given particle.

Answer to Problem 1CRE

Solution:

Average velocity [2, 5] = 0.954 ms^-1

Instantaneous velocity = 0.894 ms^-1

Explanation of Solution

Given:

$\begin{array}{l}s(t)=\sqrt{t^2+1}\\ t\in [2,5]\end{array}$

Formula used:

$\begin{array}{l}\text{Average velocity = }\frac{\text{Displacement Change}}{\text{Time Change}}\\ \text{Instantaneous velocity = }\frac{d\text{(Displacement)}}{d\text{(Time)}}\end{array}$

Calculation:

$\text{Average velocity = }\frac{\text{Displacement Change}}{\text{Time Change}}$

Displacement at t = 2

$\begin{array}{l}S(2)=\sqrt{2^2+1}=\sqrt{5}\\ S(5)=\sqrt{5^2+1}=\sqrt{26}\\ \text{Average velocity = }\frac{\text{Displacement Change}}{\text{Time Change}}\\ \text{Average velocity }=\frac{\sqrt{26}-\sqrt{5}}{5-2}=\frac{5.09-2.23}{3}=\frac{2.86}{3}=0.954m{s}^{-1}\\ \text{Instantaneous velocity}=\frac{d(\sqrt{t^2+1})}{dt}\\ \text{Instantaneous velocity}=\frac{1}{2}{(}^{t^2}(2t)=\frac{2t}{2\sqrt{(t^2+1)}}\\ \text{Velocity}(t=2)=\frac{2\times 2}{2\sqrt{(2^2+1)}}=\frac{4}{2(\sqrt{5})}=\frac{2}{\sqrt{5}}=0.894m{s}^{-1}\end{array}$

Conclusion:

To find the average velocity we can input values in the function and calculate the value. But when finding the instantaneous velocity we have to differentiate the displacement function and input the values.