Look at the graphs $(a)$ to $(d)$ carefully and state,with reasons,which of these cannot possibly represent one-dimensional motion of a particle.
(ALL OF THE ABOVE) The given $x-t$ graph,shown in $(a)$,does not represent one-dimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time.
The given $v-t$ graph,shown in $(b)$,does not represent one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time.
The given speed-time graph,shown in $(c)$,does not represent one-dimensional motion of the particle. This is because speed is a scalar quantity and cannot be negative.
The given total path length-time graph,shown in $(d)$,does not represent one-dimensional motion of the particle. This is because the total path length (distance) travelled by a particle can never decrease with time.
The given $v-t$ graph,shown in $(b)$,does not represent one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time.
The given speed-time graph,shown in $(c)$,does not represent one-dimensional motion of the particle. This is because speed is a scalar quantity and cannot be negative.
The given total path length-time graph,shown in $(d)$,does not represent one-dimensional motion of the particle. This is because the total path length (distance) travelled by a particle can never decrease with time.
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Similar Questions
$A$ car travels from $A$ to $B$ (without changing direction). During the first part of the journey,its average speed is $V_1$,and for the second part of the journey,its average speed is $V_2$. The ratio of the path length of the second part to the path length of the first part is $\sqrt{\frac{V_2}{V_1}}$. In this case,the average speed of the total path is the ....... of the average speeds of both parts. Choose the correct option.
Difficult
View SolutionAll the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.
$A$ motorbike starts from rest, attains a velocity of $10 \,m/s$ with an acceleration of $0.5 \,m/s^2$, travels $10 \,km$ with this uniform velocity, and then comes to a halt with a uniform deceleration of $0.2 \,m/s^2$. The total time of travel is: (in $\,s$)
$A$ man walks on a straight road from his home to a market $2.5 \; km$ away with a speed of $5 \; km \; h^{-1}$. Finding the market closed,he instantly turns and walks back home with a speed of $7.5 \; km \; h^{-1}$. What is the average speed of the man over the interval of time $0$ to $40 \; min$?
Difficult
View SolutionIn the $s-t$ equation $(s=10+20t-5t^2)$,match the following columns.
Column $I$ Column $II$ $(A)$ Distance travelled in $3\,s$ $(p)$ $-20$ units $(B)$ Displacement in $1\,s$ $(q)$ $15$ units $(C)$ Initial acceleration $(r)$ $25$ units $(D)$ Velocity at $4\,s$ $(s)$ $-10$ units
| Column $I$ | Column $II$ |
| $(A)$ Distance travelled in $3\,s$ | $(p)$ $-20$ units |
| $(B)$ Displacement in $1\,s$ | $(q)$ $15$ units |
| $(C)$ Initial acceleration | $(r)$ $25$ units |
| $(D)$ Velocity at $4\,s$ | $(s)$ $-10$ units |
Medium
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