Interpret the velocity–time graph

From the graph, the motion is piecewise linear over $0$ to $10\,\text{s}$:

• From $t=0$ to $t=4\,\text{s}$, velocity is positive and forms a triangle above the time axis.
• From $t=4$ to $t=10\,\text{s}$, velocity is negative and forms a triangle below the time axis.

Displacement is the signed area under the $v$–$t$ graph, while distance is the sum of magnitudes of these areas.

Area from $0$ to $4\,\text{s}$

The graph shows a triangle of base $4\,\text{s}$ and height $2\,\text{m/s}$.

$A_1=\frac{1}{2}\times 4 \times 2=4\,\text{m}$

Since it is above the axis, contribution to displacement is $+4\,\text{m}$.

Area from $4$ to $10\,\text{s}$

The graph shows a triangle below the axis with base $6\,\text{s}$ and height $4\,\text{m/s}$.

$A_2=\frac{1}{2}\times 6 \times 4=12\,\text{m}$

Since it is below the axis, contribution to displacement is $-12\,\text{m}$.

Calculate displacement

$\text{Displacement}=A_1-A_2=4-12=-8\,\text{m}$

Its magnitude is $8\,\text{m}$.

Calculate distance travelled

$\text{Distance}=|A_1|+|A_2|=4+12=16\,\text{m}$

Required ratio

$\frac{\text{displacement}}{\text{distance}}=\frac{8}{16}=\frac{1}{2}$

Hence,

$\text{displacement} : \text{distance} = 1:2$

Correct option

Option D is correct.

Comparison with stored answer

The stored correct answer is B: $1:3$, but the ratio obtained from the graph is $1:2$.

So the stored answer appears to be incorrect.