Solving for Velocity over Time: A Calculus Challenge

[Radarithm]

Homework Statement

$$F=-mg-m\alpha v$$$$F=m\frac{dv}{dt} =$$
$$\int_0^v \frac{dv'}{g+\alpha v'} = -\int_0^t dt'$$
This equals:
$$\ln \frac{1+\alpha v}{g} = -\alpha t$$

I don't understand why it does; I keep getting an incorrect answer. Can anyone tell me what I'm doing wrong?

Homework Equations

The Attempt at a Solution

let $$u=g+\alpha v'$$$$du=g+vdv$$ so $du-g=vdv$
$$-g+\int_0^v udu = \ln (g+\alpha v)-g$$
This has to do with getting the velocity as a function of time; I don't know if it should be in the physics section or not but this is a problem I'm having with calculus.

 

[Tanya Sharma]

Homework Statement

$$F=mg-m\alpha v$$$$F=m\frac{dv}{dt} =$$
$$\int_0^v \frac{dv'}{g+\alpha v'} = -\int_0^t dt'$$
This equals:
$$\ln \frac{1+\alpha v}{g} = -\alpha t$$

I don't understand why it does; I keep getting an incorrect answer. Can anyone tell me what I'm doing wrong?

Homework Equations

The Attempt at a Solution

let $$u=g+\alpha v'$$$$du=g+vdv$$ so $du-g=vdv$
$$-g+\int_0^v udu = \ln (g+\alpha v)-g$$
This has to do with getting the velocity as a function of time; I don't know if it should be in the physics section or not but this is a problem I'm having with calculus.

Is $F=mg-m\alpha v$ or $F=mg+m\alpha v$ ?

 

[Radarithm]

Is $F=mg-m\alpha v$ or $F=mg+m\alpha v$ ?

$$F=-mg-m\alpha$$

 

[HallsofIvy]

Homework Statement

$$F=-mg-m\alpha v$$$$F=m\frac{dv}{dt} =$$
$$\int_0^v \frac{dv'}{g+\alpha v'} = -\int_0^t dt'$$
This equals:
$$\ln \frac{1+\alpha v}{g} = -\alpha t$$

I don't understand why it does; I keep getting an incorrect answer. Can anyone tell me what I'm doing wrong?

Homework Equations

The Attempt at a Solution

let $$u=g+\alpha v'$$$$du=g+vdv$$ so $du-g=vdv$

This makes no sense. You chose to use v' as the variable of integration. Why does it become v when you find du? Also g is a constant so its derivative is 0. If $u= g+ \alpha v'$ then $du= \alpha dv'$ so that $dv'= du/\alpha$
$$\int_0^v \frac{dv'}{g+ \alpha v'}= \frac{1}{\alpha}\int_g^{g+\alpha v} \frac{du}{u}$$

$$-g+\int_0^v udu = \ln (g+\alpha v)-g$$
This has to do with getting the velocity as a function of time; I don't know if it should be in the physics section or not but this is a problem I'm having with calculus.

 

[Radarithm]

This makes no sense. You chose to use v' as the variable of integration. Why does it become v when you find du? Also g is a constant so its derivative is 0. If $u= g+ \alpha v'$ then $du= \alpha dv'$ so that $dv'= du/\alpha$
$$\int_0^v \frac{dv'}{g+ \alpha v'}= \frac{1}{\alpha}\int_g^{g+\alpha v} \frac{du}{u}$$

Why do the limits of integration change from $0 \to v(t)$ to $g \to g+\alpha v$ ?

 

[Tanya Sharma]

Why do the limits of integration change from $0 \to v(t)$ to $g \to g+\alpha v$ ?

Because the variable of integration has changed from v' to u .

Use the relation between v' and u (u=g+αv') to find the corresponding new limits .

When v' = 0 → u = g (lower limit)
When v' = v → u = g+αv (upper limit)

 

[HallsofIvy]

Because you changed variables from v' to $u= g+ \alpha v'$. When v'= 0, $u= g+ \alpha(0)= g$ and when v'= v, $u= g+ \alpha v$.

 

[Radarithm]

edit: double posted. sorry

 

[Radarithm]

Because the variable of integration has changed from v' to u .

Use the relation between v' and u (u=g+αv') to find the corresponding new limits .

When v' = 0 → u = g (lower limit)
When v' = v → u = g+αv (upper limit)

Because you changed variables from v' to $u= g+ \alpha v'$. When v'= 0, $u= g+ \alpha(0)= g$ and when v'= v, $u= g+ \alpha v$.

Thanks for the help guys. I appreciate it.