Help in Differential Equations - find velocity at t seconds...

Question

Hello Everyone I am stuck where I am , Would like to know if I'm going the write path, and any hints on how to proceed would be greatly appreciated !!

Question:

A boat carrying 7 people is being towed at $5 \; m/s$. the combined mass is $800 \; Kg$. The rope is suddenly cast off and immediately the riders begin to row in the direction of motion exerting a force of $2,400 \; N$. Assuming the resistance force has magnitude $1600|v| + 800v^2$, find the velocity $t$ seconds after the rope was cast off.

Where I left off:

$m=800\;Kg$

$F=ma$

$F=1600v + 800v^2 +2400 $

$ma=1600v + 800v^2 +2400$

$800a=1600v+800v^2+2400$

$800 \left(\dfrac{d^2v}{dt^2}\right) = 1600v + 800v^2 +2400$

$\dfrac{d^2v}{dt^2} = \dfrac{1600}{800}v+ \dfrac{800v^2} {800} + \dfrac{2400}{800}$

$\dfrac{d^2v}{dt^2} = 2v+ v^2 +3$

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Answer 1

You have: $$ a=\dfrac{dv}{dt}=\dfrac{F-R}{m} $$ where: $ F = 2400\; N$, $R= (1600|v|+800 v^2) \; N$ and $m=800\;Kg$. So you have to solve the equation: $$ \dfrac{dv}{dt}=-v^2-2v+3 $$ With the inital condition $v_0= 5\; m/s$

Answer 2

If you have an expression for acceleration as a function of speed $a(v)$ then the following holds true:

• Time to change velocities from $v_1$ to $v_2$ $$ \Delta t = \int \limits_{v_1}^{v_2} \frac{1}{a(v)}\,{\rm d}v$$
• Distance traveled during this time $$ \Delta x = \int \limits_{v_1}^{v_2} \frac{v}{a(v)}\,{\rm d}v$$

So for you with $a(v) = C_1 v + C_2 v^2$ you have

$$ t = \int \limits_{v_1}^{v_2} \frac{1}{C_1 v + C_2 v^2}\,{\rm d}v = \frac{1}{C_1} \ln \left( \frac{v_2 (C_1+C_2 v_1)}{v_1 (C_1+C_2 v_2) } \right)$$