Question

The one-dimensional motion of a point particle is shown in the figure. Select the correct statement:

• A. The total displacement of the particle is zero
• B. The maximum acceleration of the particle is $1/2$ ms$^{-2}$
• C. The total distance travelled by the particle at the end of 10 s is 100 m
• D. At the $5^{th}$ second, the acceleration of the particle is 2 ms$^{-2}$
• E.

Hint:

To find displacement from a $v-t$ graph, always subtract the area below the time axis from the area above it. If they are equal, the particle has returned to its origin.

Answer 1

The Correct Option is B

Concept:
In a velocity-time ($v-t$) graph: % Note: The axis label in the image provided is "Speed (ms^{-1})", but the motion extends below zero. % Physically, this is a velocity-time graph showing a change in direction. The solution correctly treats it as a v-t graph. [itemsep=8pt]
• Displacement: This is the algebraic sum of the areas under the curve (Areas above the $x$-axis are positive; areas below are negative).
• Distance: This is the sum of the magnitudes of all areas (All areas are treated as positive).

Step 1: Calculating Areas from the Graph.
The graph shown above contains two distinct triangular regions: [label=\alph*), itemsep=10pt, topsep=10pt]
• Area from 0s to 10s ($A_1$):
The base is 10s, and the height reaches a peak velocity of 10 ms$^{-1}$ at $t=5$s. This forms a large triangle from 0 to 10s.
$A_1 = \text{Area of large triangle} = \frac{1}{2} \times \text{base} \times \text{height}$
$A_1 = \frac{1}{2} \times 10 \times 10 = +50$ m.
• Area from 10s to 20s ($A_2$):
The base is $(20 - 10) = 10$s, and the curve goes below the axis to a height of $-10$ ms$^{-1}$ (reached at $t=15$s).
$A_2 = \frac{1}{2} \times \text{base} \times \text{negative\_height}$
$A_2 = \frac{1}{2} \times 10 \times (-10) = -50$ m.

Step 2: Determining Total Displacement.
Total Displacement = $A_1 + A_2$
Total Displacement = $50 \text{ m} + (-50 \text{ m}) = 0$ m.
[8pt] Since the total displacement is zero, statement (B) is correct.