(Air Resistance) Distance with respect to time

[BitterX]

Homework Statement

a ball with mass $m$ is thrown up from the ground with a velocity $v_0$

the force due to Air resistance is: $F=-bv$
(there's also gravity)

find the velocity and distance with respect to time

The Attempt at a Solution

I think I got the velocity:
$v=e^{-\frac{b}{m}t}(\frac{m}{b}g+v_0)-\frac{m}{b}g$

the units are okay and so does t=0 (v=v₀)

I have a problem with the distance, I integrate:

$x=\int^{t}_{0}vdt=\int^{t}_{0}(e^{-\frac{b}{m}t}(\frac{m}{b}g+v_0)-\frac{m}{b}g) = -e^{-\frac{b}{m}t}(\frac{m}{b}g+v_0)\frac{m}{b}-\frac{m}{b}gt$

so the units are okay but if t=0 $x=-(\frac{m}{b}g+v_0)\frac{m}{b}-\frac{m}{b}gt$

and it should be 0 (or so I think)
maybe the constant of integration should be the expression I got?
I'm really confused.

Thanks in advance!

 

[collinsmark]

Homework Statement

a ball with mass $m$ is thrown up from the ground with a velocity $v_0$

the force due to Air resistance is: $F=-bv$
(there's also gravity)

find the velocity and distance with respect to time

The Attempt at a Solution

I think I got the velocity:
$v=e^{-\frac{b}{m}t}(\frac{m}{b}g+v_0)-\frac{m}{b}g$

the units are okay and so does t=0 (v=v₀)

That looks right to me, so far.

I have a problem with the distance, I integrate:

$x=\int^{t}_{0}vdt=\int^{t}_{0}(e^{-\frac{b}{m}t}(\frac{m}{b}g+v_0)-\frac{m}{b}g)$

I'm sure as you know, there are two ways to evaluate an integral. The first way is to find the anti-derivative, and then tack on an arbitrary constant at the end. Then use your initial conditions to solve for the arbitrary constant.

The other way, which you have started here, is to evaluate the definite integral, and apply the initial conditions as your limits of integration. For example, if F(t) is the anti-derivative of f(t), then

$$F(b) - F(a) = \int_a^b f(t)dt$$

Since you've already started with the definite integral method, let's go with that.

$= -e^{-\frac{b}{m}t}(\frac{m}{b}g+v_0)\frac{m}{b}-\frac{m}{b}gt$

Okay, that looks like the correct anti-derivative to me, but you forgot to evaluate it over the limits from 0 to t.

Plugging in t = t is easy enough, but notice that the antiderivative is not 0, when evaluated at t = 0. This needs to be subtracted off. Remember,

$$F(b) {\color{red} {- F(a)}} = \int_{\color{red}{a}}^b f(t)dt$$

 

[BitterX]

Wow, I don't know how I missed it.

I plugged t=0 only in -(mg/b)t (so it was 0) and totally ignored the first part :/

Thank you!