Integrating to find distance and time - difficult

In summary, the conversation discusses a problem where a box with a parachute is dropped from a plane and the equations for velocity in terms of time are given. The variables p and z represent the drag coefficient in the x and y directions respectively, and θ is the constant angle between the ground and the box. The conversation also raises questions about the initial velocity and finding the time it takes for the box to hit the ground using integration. The solution is found by setting the initial velocity in the y direction to zero and solving for dy/dt.
  • #1
1234john
1
0
I have been looking for problems to work out for practice and came across one I have no idea how to even start about.

Say a box is dropped from a plane, with a parachute attached to it. If we are given two equations for velocity in terms of time, such as

Vx(t)=Vo cos(θ) e^(-pt) and Vy(t)=Vo sin(θ) - ge^(-z(t+6)) -16

If p is the drag coefficient in the x and z is the drag coefficient in the y.
θ would be the angle between the ground and the box.

Would θ not be a constant though since the angle will always be constant, even if the box follows a parabolic path?

And say the plane was flying level when the box was dropped... Wouldn't this mean that there is only an initial velocity in the x direction at whatever speed the plane was flying?

For practice, I made up p=0.061 and z=0.213

My whole dilemma is how I would find the time it takes for the box to hit the ground.

I'm fairly sure I'd have to take the integral which would give me distance. Plugging in zero for Vo in the y direction would cancel out the first part of the equation. θ would remain constant which would take out the sin and cos part of each equation and make it a constant.

I guess from there I am lost. My above assumptions could be wrong, however.

Please let me know! No rush on answering this as its just for my own personal practice and "fun".
 
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  • #2
as Vo is y direction is zero
Vy(t)=-g exp(-z(t+6))-16

Vy=dy/dt, it makes
dy= (-g exp(-z(t+6))-16 )dt

and you get what you want
 

FAQ: Integrating to find distance and time - difficult

How do I integrate to find distance and time?

Integrating to find distance and time involves using calculus to find the area under a curve that represents the velocity of an object. This area represents the distance traveled by the object over a given time period.

Why is integrating to find distance and time considered difficult?

Integrating to find distance and time can be difficult because it involves understanding the concept of integration and using complex formulas to calculate the area under a curve. It also requires a solid understanding of calculus and its applications.

What are the steps involved in integrating to find distance and time?

The first step is to graph the velocity function of the object over the given time period. Then, determine the limits of integration, which represent the starting and ending points on the graph. Next, use the integration formula to calculate the area under the curve. Finally, interpret the area as the distance traveled by the object.

Can integrating to find distance and time be used for any type of motion?

Yes, integrating to find distance and time can be used for any type of motion as long as the velocity function is known. This includes both constant and variable acceleration.

Are there any limitations or assumptions when using integration to find distance and time?

One limitation is that integration assumes that the velocity function is continuous and smooth. In real-world scenarios, this may not always be the case. Additionally, integration does not take into account external factors such as friction or air resistance, which can affect the actual distance traveled by an object.

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