A particle executes the motion described by x(t) = x0(1  eγt); t  0x0 > 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) 
Hint: Particle starts its motion at t = 0.
Step 1: Put t = 0 in x(t) to calculate initial position of the particle.
Given,

x(t) = x0(1  eγt)

When t = 0;  x(t) = x0(1e0) = x0(11) = 0.

Step 2: Calculate the velocity, v(t) = dxdt

v(t) = dx(t)dt = x0γeγt.

Step 3: Calculate initial velocity by putting t = 0

V(0) = x0γe0 = x0γ.

 

(b)Hint: If f is increasing dfdt > 0, if decreasing dfdt < 0.

Step 1: Check weather f is increasing or decreasing

For x(t)

As here dxdt = x0γeγt > 0 

i.e., x is increasing

For velocity, v(t)

dv(t)dt = -x0γ2e-γt < 0

i.e., v(t) is decreasing 

For acceleration, a(t)

da(t)dt = x0γ3e-γt >0

i.e., a(t) is increasing

Step 2: Calculate maximum and minimum values

For increasing, the maximum value is at t =  and minimum value is at t = 0,

For decreasing, the maximum value is at t = 0 and the minimum value is at t = .

For x(t)

As it is increasing

So,

x(t) is maximum when t=; x(t)]max=x0x(t) is minimum when t=0; x(t)]min=0v(t) is decreasingSo,v(t) is maximum when t=0; v(0)=x0γv(t) is minimum when t=; v()=0a(t) is increasing So,a(t) is maximum when t=; a()=0a(t) is minimum when t=0; a(0)=x0γ2