A particle executes the motion described by $x(t) = x_0 (1 - e^{-\gamma t} )$ ; જ્યાં $t\, \geqslant \,0\,,\,{x_0}\, > \,0$.
$(a)$ Where does the particle start and with what velocity ?
$(b)$ Find maximum and minimum values of $x(t),\, v(t)$ $a(t)$. Show that $x(t)$ and $a(t)$ increase with time and $v(t)$ decreases with time.
A particle executes the motion described by $x(t) = x_0 (1 - e^{-\gamma t} )$ ; જ્યાં $t\, \geqslant \,0\,,\,{x_0}\, > \,0$.
$(a)$ Where does the particle start and with what velocity ?
$(b)$ Find maximum and minimum values of $x(t),\, v(t)$ $a(t)$. Show that $x(t)$ and $a(t)$ increase with time and $v(t)$ decreases with time.
Given, $\quad x(t)=x_{0}\left(1-e^{-\gamma t}\right)$
$v(t)=\frac{d x(t)}{d t}=x_{0} \gamma e^{-\gamma t}$
$a(t)=\frac{d v(t)}{d t}=-x_{0} \gamma^{2} e^{-\gamma t}$
$(a)$ When $t=0 ; x(t)=x_{0}\left(1-e^{-0}\right)=x_{0}(1-1)=0$
$v(t)=x_{0} \gamma e^{-0}=x_{0} \gamma(1)=\gamma x_{0}$
$(b)$ $x(t)$ is maximum when $\mathrm{t}=\infty$
$[x(t)]_{\max }=x_{0}(1-0)=x_{0}$
$x(t)$ is minimum when $\mathrm{t}=0$
$[x(t)]_{\min }=0$
$v(t)$ is maximum when $t=0 ; v(0)=x_{0} \gamma$
$v(t)$ is minimum when $t=\infty ; v(\infty)=0$
$a(t)$ is maximum when $t=\infty ; a(\infty)=0$
$a(t)$ is minimum when $\mathrm{t}=0 ; a(0)=-x_{0} \gamma^{2}$
Similar Questions
The acceleration-time graph for a particle moving along $x$-axis is shown in figure. If the initial velocity of particle is $-5 \,m / s$, the velocity at $t=8 \,s$ is ....... $m / s$
The acceleration-time graph for a particle moving along $x$-axis is shown in figure. If the initial velocity of particle is $-5 \,m / s$, the velocity at $t=8 \,s$ is ....... $m / s$
If velocity of particle moving along $x-$ axis is given as $v = k\sqrt x $ . Then ($a$ is acceleration)
If velocity of particle moving along $x-$ axis is given as $v = k\sqrt x $ . Then ($a$ is acceleration)
$Assertion$ : Retardation is directly opposite to the velocity.
$Reason$ : Retardation is equal to the time rate of decrease of speed.
$Assertion$ : Retardation is directly opposite to the velocity.
$Reason$ : Retardation is equal to the time rate of decrease of speed.
- [AIIMS 2002]
The displacement of a body is given to be proportional to the cube of time elapsed. The magnitude of the acceleration of the body is
A monkey climbs up a slippery pole for $3$ and subsequently slips for $3$. Its velocity at time $t$ is given by $v (t) = 2t \,(3s -t)$ ; $0 < t < 3$ and $v(t) =\,-\, (t -3)\,(6 -t)$ ; $3 < t < 6$ $s$ in $m/s$. It repeats this cycle till it reaches the height of $20\, m$.
$(a)$ At what time is its velocity maximum ?
$(b)$ At what time is its average velocity maximum ?
$(c)$ At what time is its acceleration maximum in magnitude ?
$(d)$ How many cycles (counting fractions) are required to reach the top ?
A monkey climbs up a slippery pole for $3$ and subsequently slips for $3$. Its velocity at time $t$ is given by $v (t) = 2t \,(3s -t)$ ; $0 < t < 3$ and $v(t) =\,-\, (t -3)\,(6 -t)$ ; $3 < t < 6$ $s$ in $m/s$. It repeats this cycle till it reaches the height of $20\, m$.
$(a)$ At what time is its velocity maximum ?
$(b)$ At what time is its average velocity maximum ?
$(c)$ At what time is its acceleration maximum in magnitude ?
$(d)$ How many cycles (counting fractions) are required to reach the top ?