Distance from a velocity–time graph

The distance covered in a time interval is the area under the speed–time graph.

Since only a velocity–time graph is mentioned, we interpret the graph as lying above the time axis in the interval $0$ to $4\,\text{s}$, so distance equals the area under the $v$–$t$ graph.

Read the graph in parts

From the given graph, the motion from $t=0$ to $t=4\,\text{s}$ can be split into simple geometric regions:

• From $t=0$ to $t=2\,\text{s}$: a triangle of base $2\,\text{s}$ and height $10\,\text{m/s}$
• From $t=2$ to $t=3\,\text{s}$: a rectangle of width $1\,\text{s}$ and height $10\,\text{m/s}$
• From $t=3$ to $t=4\,\text{s}$: a triangle of base $1\,\text{s}$ and height $10\,\text{m/s}$

Calculate each area

(i) First triangle: $A_1 = \frac{1}{2}\times 2 \times 10 = 10\,\text{m}$

(ii) Rectangle: $A_2 = 1 \times 10 = 10\,\text{m}$

(iii) Second triangle: $A_3 = \frac{1}{2}\times 1 \times 10 = 5\,\text{m}$

Total distance

$\text{Distance} = A_1 + A_2 + A_3 = 10 + 10 + 5 = 25\,\text{m}$

Compare with options

The computed value is $25\,\text{m}$, which is not present in the listed options.

Comparison with stored answer

The stored correct answer is B: $30\,\text{m}$, but from the area calculation the answer comes out to be $25\,\text{m}$.

So I do not agree with the stored answer, assuming the graph corresponds to the usual piecewise-linear shape described above.