Question

For positive constants k and g, the velocity, v, of a particle of mass m at time t is given by v= mg/k(1−e⁻ᵏᵗ/ᵐ). At what rate is the velocity is changing at time 0 ? At t=4 ? What do your answers tell you about the motion? At what rate is the velocity changing at time 0? rate =___ At what rate is it changing at t=4 ? rate =___

Answer 1

The rate at which the velocity is changing at time 0 is mgk. The rate at which the velocity is changing at t = 4 is

mgke
`^{(-4)`. These answers tell us about the motion of the particle.

Step-by-step explanation:

The rate at which the velocity is changing at time 0 can be found by taking the derivative of the velocity equation with respect to time and evaluating it at t = 0. Taking the derivative of v with respect to t, we get dv/dt = mgke
`^{(-kt/m)`. Evaluating this expression at t = 0, we get dv/dt = mgk. Therefore, the rate at which the velocity is changing at time 0 is mgk.

At t = 4, we can once again take the derivative of the velocity equation with respect to time and evaluate it at t = 4. Taking the derivative of v with respect to t, we get dv/dt = mgke
`^{(-4k/m)`. Evaluating this expression at t = 4, we get dv/dt = mgke
`^{(-4)`. Therefore, the rate at which the velocity is changing at t = 4 is mgke
`^{(-4)`.

These answers tell us about the motion of the particle. At time 0, the rate at which the velocity is changing is mgk, indicating that the acceleration of the particle is mgk. At t = 4, the rate at which the velocity is changing is mgke
`^{(-4)`, which is smaller than mgk. This means that the acceleration of the particle is decreasing over time.