JEE MAIN - Physics (2023 - 11th April Morning Shift - No. 15)
From the $$\mathrm{v}-t$$ graph shown, the ratio of distance to displacement in $$25 \mathrm{~s}$$ of motion is:
_11th_April_Morning_Shift_en_15_1.png)
1
$$\frac{3}{5}$$
$$\frac{1}{2}$$
$$\frac{5}{3}$$
Explanation
Area under the graph from $t=0$ to $t=20 \mathrm{sec}=200 \mathrm{~m}$
Area under the graph from $\mathrm{t}=20$ to $t=25 \mathrm{sec}=50 \mathrm{~m}$
So distance covered $=(200+50) \mathrm{m}=250 \mathrm{~m}$
Displacement $=(200-50) \mathrm{m}=150 \mathrm{~m}$
$$ \therefore $$ $$ \frac{\text { distance }}{\text { displacement }}=\frac{250}{150}=\left(\frac{5}{3}\right) $$
Area under the graph from $\mathrm{t}=20$ to $t=25 \mathrm{sec}=50 \mathrm{~m}$
So distance covered $=(200+50) \mathrm{m}=250 \mathrm{~m}$
Displacement $=(200-50) \mathrm{m}=150 \mathrm{~m}$
$$ \therefore $$ $$ \frac{\text { distance }}{\text { displacement }}=\frac{250}{150}=\left(\frac{5}{3}\right) $$
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