The position-time graph of a particle moving | vedclass

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(B) The correct option is $(B)$.$(i)$ From $t=0 \, s$ to $t=2 \, s$,the position $x=0$,so the velocity $v = \frac{dx}{dt} = 0$.$(ii)$ From $t=2 \, s$

Vedclass · Accepted Answer

Velocity of the particle is opposite to acceleration between $t=2 \, s$ to $t=5 \, s$.

Vedclass · Answer

(B) The correct option is $(B)$.$(i)$ From $t=0 \, s$ to $t=2 \, s$,the position $x=0$,so the velocity $v = \frac{dx}{dt} = 0$.$(ii)$ From $t=2 \, s$ to $t=5 \, s$,the slope of the $x-t$ graph is positive (velocity $v > 0$),but the graph is concave downwards,which implies acceleration $a $(iii)$ From $t=5 \, s$ to $t=8 \, s$,the slope of the $x-t$ graph is negative (velocity $v < 0$),and the graph is concave downwards (acceleration $a < 0$). Since both $v$ and $a$ are negative,they are in the same direction.

The position-time graph of a particle moving along a straight line is shown,which is in the form of a semicircle starting from $t=2 \, s$ to $t=8 \, s$. Select the correct statement.

Similar Questions

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The position-time graph for a body moving along a straight line between $O$ and $A$ is shown in the figure. During its motion between $O$ and $A$,how many times does the body come to rest?

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