Compare the speed of the three particles $A, B$ and $C$ as per the distance-time graph shown below.

v_{A} > v_{B} > v_{C}

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v_{A} > v_{C} > v_{B}

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v_{B} > v_{A} > v_{C}

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v_{C} > v_{A} > v_{B}

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Solution

The correct option is A $v_{A} > v_{B} > v_{C}$
Given distance-time graph as

As we know that speed of the particle is the slope of the distance-time curve
So, speed of particle $A$ is given by
$v_{A} = \frac{25 - 0}{2 - 0} = 12.5 m/s$
similarly, speed of particle $B$ is given by
$v_{B} = \frac{25 - 0}{6 - 0} \approx 4 m/s$
Also, speed of particle $C$ is given by
$v_{C} = \frac{10 - 0}{7 - 0} \approx 1.4 m/s$
So, we can conclude here that $v_{A} > v_{B} > v_{C}$

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Compare the speed of the three particles A,B and C as per the distance-time graph shown below.

Compare the speed of the three particles $A, B$ and $C$ as per the distance-time graph shown below.