How Do You Calculate the Speed of a Car Based on the Differentiation Equation?

[markosheehan]

a car starts from rest. when it is at a distance s from its starting point its speed is v and its acceleration is 25v+v³. show that dv=(25+v²)ds and find (correct to 2 decimal places) its speed when s=.01
this is how i went about it
dv/dt= acceleration dv/dt=dv/ds×ds/dt=v×dv/ds

v dv/ds=25v+v³
dv/ds+25+v²
dv=(25+v²)ds i can get this but i do not understand how to get speed when s=.01

 

[I like Serena]

a car starts from rest. when it is at a distance s from its starting point its speed is v and its acceleration is 25v+v³. show that dv=(25+v²)ds and find (correct to 2 decimal places) its speed when s=.01
this is how i went about it
dv/dt= acceleration dv/dt=dv/ds×ds/dt=v×dv/ds

v dv/ds=25v+v³
dv/ds+25+v²
dv=(25+v²)ds i can get this but i do not understand how to get speed when s=.01

Hi markosheehan! Welcome to MHB! ;)

The Taylor expansion of $v(s)$ around $s=0$ gives us:
$$v(s) = v(0) + v'(0) s + \frac 12 v''(\theta s) s^2$$
where $\theta$ is some number such that $0\le \theta \le 1$.

That is:
$$v(s) \approx v(0) + v'(0) s$$
with an error of:
$$e = \frac 12 v''(\theta s) s^2 \le \frac 12 v''(s) s^2$$

We have:
$$v(0) = 0, \quad v'(0) = 25+0^2, \quad v''(s) = 2v$$
So:
$$v(s) \approx 0 + (25 + 0^2) s = 25s$$
with an error of:
$$e = \frac 12 v''(\theta s) s^2 = \frac 12 \cdot 2v(\theta s) \cdot s^2 \le v(s)s^2 \approx 25s\cdot s^2 = 25s^3$$

 

[HallsofIvy]

Your differential equation, $$dv= (v^2+ 25)ds$$ is the same as $$\frac{dv}{v^2+ 25}= ds$$ and, integrating both sides, $$\frac{1}{5}arctan(5v)= s+ C$$ or $$v= \frac{1}{5}tan(5s+ C)$$. At "its starting point", s= 0 and v= 0 so we have $$\frac{1}{5}tan(C)= 0$$. $$tan(C)= 0$$ so C= 0. $$v= \frac{1}{5}tan(5s)$$. Set s= 0.01.