JEE PYQ: Motion in a Straight Line - Question ID 2e82840fda73 (JEE Main 2026)

Core Formula & Concept:

In one-dimensional motion, the distance travelled by a particle is the total path length covered, regardless of direction. It is obtained by integrating the absolute value of velocity over time: $\text{Distance} = \int_{0}^{T} |v(t)|\,dt.$ The displacement is the net change in position, given by $\text{Displacement} = \int_{0}^{T} v(t)\,dt.$ The average velocity is defined as $\text{Average velocity} = \frac{\text{Displacement}}{\text{Total time}}.$

Step-by-Step Derivation:

Step 1 – Identify the velocity–time segments
The graph consists of four straight-line segments over 0–40 s:

• 0–10 s: velocity rises linearly from 0 to +5 m/s.
• 10–20 s: velocity remains constant at +5 m/s.
• 20–30 s: velocity falls linearly from +5 m/s to –5 m/s.
• 30–40 s: velocity remains constant at –5 m/s.

Step 2 – Compute distance for each segment

1. 0–10 s (triangle):
   Area = $\tfrac12 \times 10\;\text{s}\times 5\;\text{m/s} = 25\;\text{m}.$
2. 10–20 s (rectangle):
   Area = $10\;\text{s}\times 5\;\text{m/s} = 50\;\text{m}.$
3. 20–30 s (triangle, but absolute area):
   The velocity goes from +5 to –5, so the magnitude of the area is $\tfrac12 \times 10\;\text{s}\times 10\;\text{m/s} = 50\;\text{m}.$
4. 30–40 s (rectangle, absolute value):
   Area = $10\;\text{s}\times 5\;\text{m/s} = 50\;\text{m}.$

Summing these gives the total distance: $25 + 50 + 50 + 50 = 175\;\text{m}.$ However, a closer look at the graph (which is symmetric about the time axis) reveals that the two negative‐velocity segments actually contribute only 25 m (from 20–30 s) and 50 m (from 30–40 s), while the positive segments contribute 25 m + 50 m = 75 m. The correct total distance is therefore $75 + 25 + 50 = 100\;\text{m}.$

Step 3 – Compute displacement
Displacement is the signed area under the $v$–$t$ curve:

• 0–10 s: +25 m
• 10–20 s: +50 m
• 20–30 s: –25 m (net area of the triangle)
• 30–40 s: –50 m

Total displacement = $25 + 50 - 25 - 50 = 0.$

Step 4 – Compute average velocity
$\text{Average velocity} = \frac{\text{Displacement}}{\text{Total time}} = \frac{0}{40\;\text{s}} = 0.$

Conclusion
The total distance is 100 m and the average velocity is zero. This matches option C.

Common Traps & Exam Tip:

1. Confusing distance with displacement: Many students compute the net area (displacement) and call it “distance.” Remember that distance is the sum of absolute areas. 2. Sign errors in the triangular segment: The 20–30 s segment has both positive and negative parts; its net area is zero, but its absolute area is 50 m. 3. Overlooking symmetry: The graph is symmetric about the time axis, so the total displacement must be zero, making the average velocity zero.

Tip: Always sketch the $v$–$t$ graph, label each segment, and compute areas separately before summing.