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:
$\left(\frac{3 v_{0}^{2}}{2 \alpha}\right)^{\frac{1}{2}}$
$\left(\frac{2 v_{0}}{3 \alpha}\right)^{\frac{1}{3}}$
$\left(\frac{2 v_{0}^{2}}{3 \alpha}\right)^{\frac{1}{2}}$
$\left(\frac{3 v_{0}^{2}}{2 \alpha}\right)^{\frac{1}{3}}$
Draw $x \to t$ graph for positive acceleration.
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 |
Equation of motion of a body is $\frac{d v}{d t}=-4 v+8$, where $v$ is the velocity in $m / s$ and $t$ is the time in second. Initial velocity of the particle was zero. Then,
A particle experiences a constant acceleration for $20\, seconds$ after starting from rest. If it travels a distance $s_1$ in the first $10\, seconds$ and distance $s_2$ in the next $10\, seconds$, then :-
Given figure shows the $x-t$ plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for $t < 0$ and on a parabolic path for $t >0$? If not, suggest a suitable physical context for this graph.