Question

A force F = Be^-ct acts on a particle whose mass is m and whose velocity is 0 at t = 0. It's terminal velocity after a long time is?

a) C/mB
b) B/mC
c) BC/m
d) -B/mC

dont spam

Answer 1

How do I solve it? A force F= B e^(-Ct) acts on a particle whose mass is m and whose velocity is 0 at t= 0 . Its terminal velocity is.

F = B*e^(-C*t)

Here, force is varying with time. So acceleration produced by the force will be;

a = (B/m)*e^(-C*t)

Where m is the mass of the body upon which the force is acting.

Terminal velocity implies the velocity when time tends to infinity.

Since, acceleration is varying, the change in velocity can be obtained by integrating acceleration with respect to time:

v - [initial velocity at (t=0)] = integral (a*dt)

v - 0 = [(-B/m*C)*e^(-C*t)] {from 0 to infinity}

Putting the values of t=0 and t=infinity;

v - 0 = [0 - (-B/m*C)*e^(-C*{t=infinity})]

v = B/m*C

Answer 2

Answer:

A force F = Be^-ct acts on a particle whose mass is m and whose velocity is 0 at t = 0. It's terminal velocity after a long time is?

a) C/mB

b) B/mC$\sqrt{}$

c) BC/m

d) -B/mC

Explanation

A force F= B e^(-Ct) acts on a particle whose mass is m and whose velocity is 0 at t= 0 . Its terminal velocity is.

F = B*e^(-C*t)

Here, force is varying with time. So acceleration produced by the force will be;

a = (B/m)*e^(-C*t)

Where m is the ma ss of the body upon which the force is acting.

Terminal velocity implies the velocity when time tends to infinity.

Since, acceleration is varying, the change in velocity can be obtained by integrating acceleration with respect to time:

v - [initial velocity at (t=0)] = integral (a*dt)

v - 0 = [(-B/m*C)*e^(-C*t)] {from 0 to infinity}

Putting the values of t=0 and t=infinity;

v - 0 = [0 - (-B/m*C)*e^(-C*{t=infinity})]

v = B/mC