homework 4

1.3. It is often the case that the frictional force on an object will increase as the object moves faster. A fortunate example of this is a parachutist; the role of the parachute is to produce a frictional force due to air drag, which is larger than would normally be the case without the parachute. The physics of air drag will be discussed in more detail in the next chapter. Here we consider a very simple example in which the frictional force depends on the velocity. Assume that the velocity of an object obeys an equation of the form
$\frac{dv}{dt}=a-bv$
where a and b are constants. You could think of a as coming from an applied force, such as gravity, while b arises from friction. Note that the frictional force is negative (we assume that b > 0), so that it opposes the motion, and that it increases in magnitude as the velocity increases. Use the Euler method to solve (1.10) for v as a function of time.- A convenient choice of parameters is a = 10 and b = 1. You should find that 1; approaches a constant value at long times; this is called the terminal velocity.
解：
$v(\Delta t)= v(0)+\frac{dv(t)}{dt}\Delta t+\frac{1}{2}\frac{d^{2}v(t)}{d^{2}t}(\Delta t)^{2}+...$
只保留一阶项有
$V(\Delta t)\approx V(0)+\frac{dv}{dt}\Delta t$
整理得
$\frac{dv(t)}{dt}=\lim_{\Delta t\rightarrow 0}\frac{v(t+\Delta t)-v(t)}{\Delta t}\approx \frac{v(t+\Delta t)-v(t)}{\Delta t}$
因为$\frac{dv}{dt}=a-bv$
所以代入上式得$v（t+\Delta t)\approx v(t)+(a-bv)\Delta t$
接下来编写代码
代码
运行并画出图像得：

初速度为0时：
初速度为20m/s时：

结论

降落伞模型中，如果速度低于10m/s,此时重力大于阻力，速度会增加，速度增加后，同时会伴随着阻力增大，此时合力变小，加速度减小，但方向仍竖直向下，所以图像上斜率逐渐变小，速度一直增加到10m/s后，此时重力等于阻力，合力为0，加速度为0，速度保持恒定，匀速下降；

同样如果速度高于10m/s,此时重力小于阻力，速度会降低，速度减小后，同时会伴随着阻力减小，此时合力变小，加速度减小，但方向仍竖直向上，保持减速，所以图像上斜率逐渐变小，直到速度减小为10m/s，此时重力等于阻力，速度保持恒定，匀速下降；