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MCQEasyJEE 2023Graphs of Motion

JEE Physics 2023 Question with Solution

The velocity-time graph of a body moving in a straight line is shown in the figure.

Velocity-time graph with time from 0 to 10 s on horizontal axis and velocity in m s^{-1} on vertical axis, showing piecewise constant segments above and below the time axis.

The ratio of displacement to distance travelled by the body in time 00 to 10 s10 \, \text{s} is:

  • A

    1:11:1

  • B

    1:41:4

  • C

    1:21:2

  • D

    1:31:3

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The required ratio is displacement to distance travelled from 00 to 10 s10 \, \text{s} using the velocity-time graph.

Find: displacementdistance\dfrac{\text{displacement}}{\text{distance}}

From the velocity-time graph, we calculate the displacement and total distance traveled.

Displacement: The net displacement is the area under the graph considering direction.

Displacement=16+(−8)+16+(−8)=16 m\text{Displacement} = 16 + (-8) + 16 + (-8) = 16 \, \text{m}

Distance: The total distance is the sum of the absolute areas under the graph.

Distance=∣16∣+∣8∣+∣16∣+∣8∣=48 m\text{Distance} = |16| + |8| + |16| + |8| = 48 \, \text{m}

Therefore,

DisplacementDistance=1648=13\frac{\text{Displacement}}{\text{Distance}} = \frac{16}{48} = \frac{1}{3}

Therefore, the ratio of displacement to distance travelled is 1:31:3. The correct option is D.

Area Interpretation

Given: A piecewise constant velocity-time graph over 00 to 10 s10 \, \text{s}.

Find: The ratio of net displacement to total distance.

Use the fact that area above the time axis contributes positive displacement, while area below the time axis contributes negative displacement. Distance uses absolute value of each area.

From the graph,

Displacement=Σarea=16−8+16−8=16 m\text{Displacement} = \Sigma \text{area} = 16 - 8 + 16 - 8 = 16 \, \text{m}Distance=Σ∣area∣=16+8+16+8=48 m\text{Distance} = \Sigma |\text{area}| = 16 + 8 + 16 + 8 = 48 \, \text{m}

Hence,

ratio=1648=13\text{ratio} = \frac{16}{48} = \frac{1}{3}

So the required ratio is 1:31:3.

Common mistakes

  • Taking distance and displacement as the same quantity. This is wrong because displacement is signed area, while distance is the sum of absolute areas. Use algebraic sum for displacement and absolute sum for distance.

  • Ignoring the portions of the graph below the time axis. This is wrong because negative velocity contributes negative displacement but still adds positively to distance. Count those sections with sign for displacement and magnitude for distance.

  • Using the listed answer key label instead of the worked value. This is wrong because the solution shows the ratio is 1:31:3, which corresponds to option D in the given options. Always map the computed value to the actual option list.

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