''The magnitude of average velocity is equal to average speed''. This statement is not always correct and not always incorrect. Explain with example.
''The magnitude of average velocity is equal to average speed''. This statement is not always correct and not always incorrect. Explain with example.
Suppose, a car takes $18 \mathrm{~s}$ to move from $\mathrm{O}(0)$ to $\mathrm{P}(360)$. Now in next $6 \mathrm{~s}$, it returns from $\mathrm{P}$ to Q (240), then by calculating average velocity and average speed the answer of this question can be explained.
For motion from $\mathrm{O}$ to $\mathrm{P}$,
$\text { Average speed } =\frac{\text { OP }}{t_{1}}=\frac{+360}{18}=20 \mathrm{~ms}^{-1}$
$\text { Average velocity } =\frac{\text { Position of } P-\text { Position of } Q}{t_{1}}$
$=\frac{360-0}{18}$
$=\frac{360}{18}=20 \mathrm{~ms}^{-1}$
Here, both are equal.
For motion from $\mathrm{P}$ to $\mathrm{Q}$,
$\text { Average speed } =\frac{\mathrm{OP}+\mathrm{PQ}}{t_{2}}=\frac{360+120}{t_{1}+6}[\because \mathrm{PQ}=360-240=120]$
$=\frac{480}{18+6}=\frac{480}{24}$
$=20 \mathrm{~ms}^{-1}$
$\text { Average velocity } =\frac{\text { Position of } \mathrm{Q}-\text { Position of } \mathrm{O}}{t_{2}}$
$=\frac{240}{24}$
$=10 \mathrm{~ms}^{-1}$
Here, both are not equal.
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