Question

Imagine a small child whose legs are half as long as her parent's legs. If her parent can walk at maximum speed V , at what maximum speed can the child walk?
a-v
b-v/2
c-2v
d-sqrt2* v
e-v/sqrt 2

Answer 1

Answer:

The maximum speed can the child walk is v/sqrt 2

Explanation:

Let the length of parent’s leg be L

Then the length of the child’s leg $=\frac{L}{2}$

Maximum Speed at which the parent walks=V

To Find :

The maximum speed at which the child walks=?

Solution:

The Frequency of the simple pendulum

$f=\frac{1}{2 \pi}(\sqrt{\frac{g}{l}})$

$\text {Speed}=\frac{\text {distance}}{\text {time}}$

Speed of the parent  

$V=L f_{p}$

$V=L\left(\frac{1}{2 \pi}(\sqrt{\frac{g}{L}})\right)$

Speed of the child

$v=\frac{L}{2}\left(\frac{1}{2 \pi}(\sqrt{\frac{g}{L / 2}})\right)$

Now,

$\frac{v}{V}=\frac{\frac{L}{2}\left(\frac{1}{2 \pi}(\sqrt{\frac{g}{L / 2}})\right)}{L\left(\frac{1}{2 \pi}(\sqrt{\frac{g}{L}})\right)}$

$\frac{v}{V}=\frac{\sqrt{2}}{2}$

$\frac{v}{V}=\frac{\sqrt{2}}{\sqrt{2} \sqrt{2}}$

$\frac{v}{V}=\frac{1}{\sqrt{2}}$

$v=\frac{V}{\sqrt{2}}$

Result:

The maximum Speed at which the child can walk is$\frac{V}{\sqrt{2}}$