Velocity of a particle is in negative direction with constant acceleration in positive direction. Then, match the following columns.
Colum $I$ | Colum $II$ |
$(A)$ Velocity-time graph | $(p)$ Slope $\rightarrow$ negative |
$(B)$ Acceleration-time graph | $(q)$ Slope $\rightarrow$ positive |
$(C)$ Displacement-time graph | $(r)$ Slope $\rightarrow$ zero |
$(s)$ $\mid$ Slope $\mid \rightarrow$ increasing | |
$(t)$ $\mid$ Slope $\mid$ $\rightarrow$ decreasing | |
$(u)$ |Slope| $\rightarrow$ constant |
$( A ) \rightarrow Q , T _{;}( B ) \rightarrow Q , S ;( C ) \rightarrow P , T$
$( A ) \rightarrow Q , U ;( B ) \rightarrow R , U ;( C ) \rightarrow P , T$
(A) $\rightarrow P , T ;( B ) \rightarrow R , U ;( C ) \rightarrow Q , S$
$( A ) \rightarrow P , T ;( B ) \rightarrow Q , U ;( C ) \rightarrow Q , T$
What does the area of $v\to t$ graph of moving object represent ?
The distance $x$ covered by a particle in one dimensional motion varies with time $t$ as $\mathrm{x}^{2}=\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c.}$ If the acceleration of the particle depends on $\mathrm{x}$ as $\mathrm{x}^{-\mathrm{n}},$ where $\mathrm{n}$ is an integer, the value of $\mathrm{n}$ is
The relation between time and distance is $t = \alpha {x^2} + \beta x$, where $\alpha $ and $\beta $ are constants. The retardation is
A particle is projected with velocity $v_{0}$ along $x-$ axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e., $ma =-\alpha x ^{2}.$ The distance at which the particle stops:
The velocity $(v)$-time $(t)$ graph for a particle moving along $x$-axis is shown in the figure. The corresponding position $(x)$ - time $(t)$ is best represented by