Question

An objects acceleration decreases exponentially with time: a(t)= a0e^−bt , where a 0 and b are constants.Assuming the object starts from rest, determine its velocity as a function of time.

I got -(a0/b)e^-bt and it's wrong. I will apreciate your help.

Answer 1

The velocity as a function of time is given by v(t) = -a0/b * e^(-bt) + a0/b.

Step-by-step explanation:

To determine the velocity as a function of time, we need to integrate the acceleration function. Given that the acceleration is given by a(t) = a0e^(-bt), where a0 and b are constants, we can integrate this to obtain the velocity function:

v(t) = ∫a(t)dt = -a0/b * e^(-bt) + C,

where C is the constant of integration. Since the object starts from rest, the initial velocity at t = 0 is 0. So we can set v(0) = 0 and solve for C:

0 = -a0/b * e^(0) + C, C = a0/b.

Therefore, the velocity as a function of time is:

v(t) = -a0/b * e^(-bt) + a0/b.

Answer 2

`v(t)=(a_o)/(b)[{1-e^(-bt)]`

Step-by-step explanation:

The acceleration of an object decreases exponentially with time as :

`a(t)=a_oe^(-bt)`

We know that, the relation between the velocity and the acceleration is given by :

`v(t)=\int\limits{a(t).dt}`

Put the value of a(t) in above equation. So,

`v(t)=\int\limits{(a_oe^(-bt)).dt}`

`v(t)=a_o\int\limits{(e^(-bt)).dt}`

`v(t)=(-a_oe^(-bt))/(b)} +c`

At t = 0, v(t) = 0

So,
`0=(-a_oe^(-b(0)))/(b)} +c`

`k=(a_o)/(b)`

So, its velocity is given by :

`v(t)=(-a_oe^(-bt))/(b)} +(a_o)/(b)`

`v(t)=(a_o)/(b)[{1-e^(-bt)]`

Hence, this is the required solution.