Question

A frog hops along a straight line Path from point 'A' to point 'B' in 10 s and then turns and hops to point 'C' in another 5 s.
Calculate the average speed and average velocity of the frog for the motion between
(a) A to B
(B) A to C (through B)​

Answer 1

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(a) For the motion from A to B:

$\displaystyle{\sf{Average\:Speed= \frac{Distance}{Time} = \frac{7-(-2)}{10}}}$

$\Rightarrow{\sf{Average\:Speed= \dfrac{9}{10} = 0.9ms^{-1}}}$

$\displaystyle{\sf{Average\:velocity= \frac{Displacement}{Time} = \dfrac{7-(-2)}{10}}}$

$\Rightarrow{\sf{Average\:velocity= \dfrac{9}{10} = 0.9ms^{-1}}}$

(b) For the motion from A to C through B :

$\displaystyle{\sf{Average\:Speed= \frac{Total\:Distance}{Total\:Time} = \dfrac{[7-(-2)]+ [3-(-2)]}{10+5}}}$

$\Rightarrow{\sf{ \dfrac{14}{15} = 0.93ms^{-1}}}$

$\displaystyle{\sf{Average\:Speed= \frac{Net\:Displacement}{Total\:Time} = \dfrac{7-3}{15}}}$

$\Rightarrow{\sf{ \dfrac{4}{15} = 0.27ms^{-1}}}$.

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