From above graph choose correct option for $t_{1}$ to $t_{2}$ where $t_{1}$ is the time at which velocity of particle = 0 $t_{2}$ is the time at which velocity of particle is negative.

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Solution

The correct option is C

Particle seems to be at rest at t = 10 sec as even before and after t = 10 sec mark the particle continues to be at 10m displacement from starting point. Thus its displacement doesn't change.

So at t = 10 sec, velocity = 0

Now for the velocity to be negative

Let's analyze the motion of the particle: till time particle moved away from starting point and reached a distance of 10m.

Till $t = t_{c}$ particle stayed there maintaining it displacement of 10m.

Now after $t_{c}$ the particle starts moving towards the starting point as its displacement is decreasing.

So particles velocity was leftwards after $t = t_{c}$

$\Rightarrow$ Particle was moving with a negative velocity in the time instance t = 17 sec, 20 sec, 30 sec etc.

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From above graph choose correct option for t1 to t2 where t1 is the time at which velocity of particle = 0 t2 is the time at which velocity of particle is negative.

From above graph choose correct option for $t_{1}$ to $t_{2}$ where $t_{1}$ is the time at which velocity of particle = 0 $t_{2}$ is the time at which velocity of particle is negative.