Differential equations question

[Spoti112]

I saw this problem and solved it but still I had one question...

Homework Statement

A rock falls through water with a continuously decreasing acceleration. Assume that the rock’s acceleration as a function of velocity has the form a_y = g - bv_y where b is a positive constant. (The +y direction is directly downward.) Prove mathematically that if the rock is released from rest at time t=0, the acceleration will depend exponentially on time according to a_y(t) = g*e^-2bt.

a_y = g - bV_y , b>0 , b=const
t=0 → V₀=0 m/s

Homework Equations

dV_y/dt = a_y
V_y = ∫a_ydt

The Attempt at a Solution

a_y = g - bV_y
dV_y/dt = g - bV_y
dV_y / (g - bV_y) = dt ⇒ t = -(1/b)*ln(g - bV_y) + c
0=-(1/b)*ln(g - bV_y) + c ⇒ c = (ln(g)) / b
⇒ bt = ln(g) - ln(a_y) ⇒ a_y(t) = g*e^-2bt

BUT... when i do it this way it doesn't i get this: (this was my first try)
V_y = ∫a_ydt = gt - bV^yt + c
t=0 → V₀=0 ⇒ c=0
V_y = gt - bV^yt ⇒ V_y * (1+bt) = gt
a_y = dV_y/dt = (d/dt) * ( gt/(1+bt) ) = g/(bt+1)² → ?

is this second answer correct?

 

[Charles Link]

I saw this problem and solved it but still I had one question...

Homework Statement

A rock falls through water with a continuously decreasing acceleration. Assume that the rock’s acceleration as a function of velocity has the form a_y = g - bv_y where b is a positive constant. (The +y direction is directly downward.) Prove mathematically that if the rock is released from rest at time t=0, the acceleration will depend exponentially on time according to a_y(t) = g*e^-2bt.

a_y = g - bV_y , b>0 , b=const
t=0 → V₀=0 m/s

Homework Equations

dV_y/dt = a_y
V_y = ∫a_ydt

The Attempt at a Solution

a_y = g - bV_y
dV_y/dt = g - bV_y
dV_y / (g - bV_y) = dt ⇒ t = -(1/b)*ln(g - bV_y) + c
0=-(1/b)*ln(g - bV_y) + c ⇒ c = (ln(g)) / b
⇒ bt = ln(g) - ln(a_y) ⇒ a_y(t) = g*e^-2bt

BUT... when i do it this way it doesn't i get this: (this was my first try)
V_y = ∫a_ydt = gt - bV^yt + c
t=0 → V₀=0 ⇒ c=0
V_y = gt - bV^yt ⇒ V_y * (1+bt) = gt
a_y = dV_y/dt = (d/dt) * ( gt/(1+bt) ) = g/(bt+1)² → ?

is this second answer correct?

Where does the 2 come from in the exponential in your first answer? I believe there should not be any 2 there. $\\$ For this problem, I like to write the equation $\frac{dv}{dt}=g-bv$ as a differential equation $\frac{dv}{dt}+bv=g$ and then find the homogeneous solution, along with the particular solution for $v$. Taking $a=\frac{dv}{dt}$ gives the final result. I got the same answer in that manner, but again, without a 2 in the exponential. $\\$ For your second method, you don't know what the function $v=v(t)$ looks like, so you can't integrate $\int bv \, dt$. The result you give for the integral is incorrect, and really undetermined, until you know what $v=v(t)$ is.

 

[Spoti112]

sorry there is no 2. I have made a mistake when i was writing this post... and i copied it everywhere...

 

[Spoti112]

For your second method, you don't know what the function v=v(t)v=v(t) v=v(t) looks like, so you can't integrate ∫bvdt∫bvdt \int bv \, dt . The result you give for the integral is incorrect, and really undetermined, until you know what v=v(t)v=v(t) v=v(t) is.

don`t i know the function from the V(t=0)=0 and the differential equation that i get from the integral Vy = ∫a_ydt (here i integrate a known function) ?
i mean aren't these two thing enough?

 

[Charles Link]

don`t i know the function from the V(t=0)=0 and the differential equation that i get from the integral Vy = ∫a_ydt (here i integrate a known function) ?
i mean aren't these two thing enough?

The statement $v=\int\limits_{0}^{t} a \, dt +v_o$ is correct, but when you put in the form $a=g-bv$ , you are now working with $v=v(t)$ in the integral. $v$ is not a constant in this integrand, and you do not know its functional form $v=v(t)$.

 

[Spoti112]

GOT it! thank you very much!