(III) The force of air resistance (drag force) on a rapidly...

(III) The force of air resistance (drag force) on a rapidly falling body such as a skydiver has the form $F_{\mathrm{D}}=-k v^{2},$ so that Newton's second law applied to such an object is $$m \frac{d v}{d t}=m g-k v^{2},$$ where the downward direction is taken to be positive. (a) Use numerical integration to estimate (within 2$\% )$ the position, speed, and acceleraton, from $t=0$ up to $t=15.0 \mathrm{s},$ for a $75-\mathrm{kg}$ skydiver who starts from rest, assuming $k=0.22 \mathrm{kg} / \mathrm{m} .$ (b) Show that the diver eventually reaches a steady speed, the terminal speed, and explain why this happens. (c) How long does it take for the skydiver to reach 99.5$\%$ of the terminal speed?

(III) The force of air resistance (drag force) on a rapidly
falling body such as a skydiver has the form $F_{\mathrm{D}}=-k v^{2},$ so
that Newton's second law applied to such an object is
$$m \frac{d v}{d t}=m g-k v^{2},$$
where the downward direction is taken to be positive.
(a) Use numerical integration to estimate (within 2$\% )$ the
position, speed, and acceleraton, from $t=0$ up to
$t=15.0 \mathrm{s},$ for a $75-\mathrm{kg}$ skydiver who starts from rest,
assuming $k=0.22 \mathrm{kg} / \mathrm{m} .$ (b) Show that the diver eventually
reaches a steady speed, the terminal speed, and explain why
this happens. (c) How long does it take for the skydiver to
reach 99.5$\%$ of the terminal speed?

Key Concepts

Terminal Velocity

Terminal velocity is the constant speed that a falling object reaches when the force of gravity is balanced by the drag force opposing its motion, resulting in zero net acceleration. The concept is vital in understanding how systems evolve over time, as the forces eventually achieve equilibrium leading to steady state motion.

Newton's Second Law

Newton’s Second Law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. This principle underlies the analysis of motion and is essential for setting up the differential equations that describe the dynamics of systems subjected to multiple forces, such as gravitational and drag forces.

Drag Force

The drag force is a resistive force that opposes an object's motion through a fluid (like air) and often depends on the velocity of the object. In many cases, including for high-speed motions, the drag force is modeled as proportional to the square of the velocity. This quadratic dependence is crucial in understanding why objects experience increasing resistance at higher speeds.

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives, representing how a physical quantity changes over time. They are used to model dynamical systems such as the motion of a falling body under gravity and drag, enabling predictions about position, velocity, and acceleration over time.

Numerical Integration

Numerical integration techniques are used to approximate the solutions of differential equations when analytical solutions are difficult or impossible to obtain. Methods such as Euler's method or Runge-Kutta methods allow us to estimate the evolution of a system’s state variables (like position and velocity) over discrete time steps while maintaining a controlled level of error.

Transcript

00:01 So for a rapidly falling body, we are considering the downward direction as positive.

00:12 So that's positive.

00:15 And then we have the drag force, negative kv squared.

00:20 And the equation we're given is m dv over d t, which is equal to m g minus kv squared.

00:31 Now, as we see here, dvdt is the acceleration, or in other words, change in velocity over time is acceleration.

00:43 So we can relate dvd with acceleration, and mass times acceleration is force.

00:47 So basically, this equation becomes f equals mg minus kv squared.

00:54 So if we consider the starting position as y equal to zero, and then as it goes on, on y increases then for the acceleration we'll have a which is equal to g minus k over m v squared and the reason for that is because f equals m over a so we divide m throughout to get the acceleration now for t equals to zero uh the displacement y at t equals zero is y's y not which is then similarly the velocity v0 is 0 and the acceleration a knot is equal to g minus k over m v squared where v is equal to 0 so that becomes 9 .8 meter per second squared because v is 0 equals 0 now if we assume that the acceleration is constant over the next interval and so y1 which is the next time interval basically when time equals to one we have y0 plus v0 delta t plus half of a 0 delta t whole squared and also for the velocity we have v1 equals to v0 plus a 0 plus a 0 0 whole squared and also for the velocity we have v1 equals to v0 plus a 0 delta t and for acceleration we have a 1 which is equal to negative g k by m p squared so sorry it's going to be positive yeah so basically we can apply this method for a time interval 1 second and get the speed at position 15 seconds so yeah i get the position of 15 seconds and then we reduce the interval to 0 .5 seconds again and then again find the speed and position to 15 seconds so basically what we're doing is we're setting delta t equals one second to find the position speed and acceleration we do the same thing with reduced interval 0 .5 seconds and find the position velocity and acceleration and see if they agree within 2 % or not now if that's not, then we use a smaller interval and then we do the same thing again.

03:55 Now, in this problem, the results for position and velocity agree within 2 % when we take delta t as 1 second and 0 .5 seconds, as you can see over here.

04:10 But the acceleration doesn't...

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