Question

A particle executes the motion described by $x\left(t\right)=x_0(1-e^{-\gamma t});t\ge 0,x_0<0.$

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.

Answer 1

Step 1:Find maximum and minimum value of x(t).

Given,

displacement $x\left(t\right)=x_0(1-e^{-\gamma t})$

Displacement of the particle will be maximum when time is maximum,
$x(t\to \infty )=x_0(1-e^{-\gamma \left(\infty \right)}=x_0(1-0\left)\right)$

$x(t\to \infty )=x_0$

As time , so minimum value of time will be $t=0$

Displacement will be minimum at time $t=0$

$x\left(0\right)=x_0(1-e^{-}0)=x_0(1-1)$

$x\left(0\right)=0$

Step 2: Find velocity of the particle. Formula used: $v=\frac{dx}{dt}$

By differentiating displacement equation with respect to the time

$v\left(t\right)=\frac{dx}{dt}$

$v\left(t\right)=x_0\gamma e^{-\gamma t}$

Step 3: Find minimum and maximum velocity of particle.

As velocity has inverse relation with time. So, Velocity of the particle will be minimum when time is maximum,

$(t\to \infty )=x_0\left(\gamma e^{-\gamma \left(\infty \right)}\right)=x_0\left(0\right)$

$v(t\to \infty )=0$


velocity will be maximum at time $t=0$

$v\left(0\right)=x_0\gamma \left(e^{-}0\right)=x_0\gamma \left(1\right)$

$v\left(0\right)=x_0\gamma$

As velocity has inverse relation with time. So, velocity will decrease with time.

Step 4: Find acceleration of the particle.

Formula used: $a=\frac{dv}{dt}$

By differentiating velocity equation with respect to the time

$a=\frac{dv}{dt}$

$a\left(t\right)=-x_0{\gamma }^{2}e-\gamma t$

Step 5: Find minimum and maximum acceleration of particle.
Hint: acceleration has inverse relation with time with negative sign. As acceleration has inverse relation with time with negative sign. So, acceleration of the particle will be maximum when time is maximum, $t\to$

$a(t\to \infty )=-x_0\left({\gamma }^2{e}^{-}\gamma \right(\infty \left)\right)=-x_0{\gamma }^2\left(0\right)$

$a(t\to \infty )=0$

acceleration will be minimum at time $t=0$

$a\left(0\right)=-x_0{\gamma }^2\left(e^{-}0\right)$

$a\left(0\right)=-x_0{\gamma }^2$

As acceleration has inverse relation with time with negative sign. So, acceleration will increase with time.

Final answer: $x_0;x_0\gamma ,0;0,-x_0{\gamma }^2$