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''The magnitude of average velocity is equal to average speed''. This statement is not always correct and not always incorrect. Explain with example.
Solution
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.
