Kinematics

JEE Main level, single correct, medium

81 of 308

Question 81 of 308

The velocity (vv) versus time (tt) plot of a particle is shown in the figure, for a time interval of 40 s40\text{ s}. The total distance travelled by the particle and the average velocity during this period are, respectively _______.

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Answer: (C) 100 m100\text{ m} and zero

The total distance travelled by the particle is equal to the sum of the magnitudes of the areas under the velocity-time graph.

Area of the first triangle (from t=0t = 0 to t=20 st = 20\text{ s}) is:
A1=12×20×5=50 mA_1 = \dfrac{1}{2} \times 20 \times 5 = 50\text{ m}

Magnitude of the area of the second triangle (from t=20t = 20 to t=40 st = 40\text{ s}) is:
∣A2∣=∣12×20×(−5)∣=50 m|A_2| = \left| \dfrac{1}{2} \times 20 \times (-5) \right| = 50\text{ m}

Total distance =A1+∣A2∣=50+50=100 m= A_1 + |A_2| = 50 + 50 = 100\text{ m}

The total displacement is the algebraic sum of the areas under the velocity-time graph.

Total displacement =A1+A2=50−50=0 m= A_1 + A_2 = 50 - 50 = 0\text{ m}

Average velocity is the ratio of total displacement to total time.

Average velocity =Total displacementTotal time=040=0= \dfrac{\text{Total displacement}}{\text{Total time}} = \dfrac{0}{40} = 0

Answer: 100 m100\text{ m} and zero

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