Question 27: Motion in 1D

Question

From the $$\mathrm{v}-t$$ graph shown, the ratio of distance to displacement in $$25 \mathrm{~s}$$ of motion is:

Accepted Answer

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 $=(200-50) \mathrm{m}=150 \mathrm{~m}$
$$ 	herefore $$ $$
\frac{	ext { distance }}{	ext { displacement }}=\frac{250}{150}=\left(\frac{5}{3}ight)
$$

Answer

1

Answer

$$\frac{3}{5}$$

Answer

$$\frac{1}{2}$$

Answer

$$\frac{5}{3}$$

Motion in 1D

From the $\mathrm{v}-t graph shown, the ratio of distance to displacement in 25 \mathrm{~s}$ of motion is:

Choose the correct answer: