How to find time and velocity in second order de.

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Dimitar Dimitrov
Dimitar Dimitrov on 16 Oct 2021
Commented: Yongjian Feng on 16 Oct 2021
Hello,I am new to MatLAb and I am very frustruated with this problem:
Basicaly, we need to find the time it takes for the parachutist to hit the ground and his final velocity. The mass is 87kg, the drag coefficient when the parachute is not openned is Cd=0.023.Also the inital value of y at time zero is 3km or 3000m.Also g is 9.8 m/s but this is widly known. Velocity here is the first derrivative. But I dont know how to get time and velocity . Can someome explain to me how it should be done? I woud appreciate it

Answers (1)

Yongjian Feng
Yongjian Feng on 16 Oct 2021
For a numerical approximation, maybe you can try this.
First some background information of Physics here. The first derivative of y is speed, and the second derivative of y is acceleration.
Now depending on what accuracy you want to get, pick a small time period (delta_t). Start with delta_t = 1 second first for example. Adjust it later.
Here is the approach.
  • For each time period, speed determines how far it drops. Acceleration determine how speed increases.
  • Assuming for a short time period, speed is constant, acceleration is constant for approximation, so we can easily compute how y increases and how speed increases.
Look at the first several seconds for example:
  1. The first second, Initial speed is 0, so it does not drop in the first second (again this is an approximation), accelaration is -9.8 m/s^2. Speed increases to -9.8m/s.
  2. The second second, now speed is -9.8m/s, so it drops 9.8m. Acceleration is (0.023/87)*(-9.8)^2 - 9.8. So the speed is acceleration*1sec-9.8m/s.
Repeat the above steps and accumlate the distance it drops. Once the accumulation reaches 3000m, you get the time and the speed.
You can see this approach, the accuracy of time is 1sec. To get to higher accuracy, you need to decrease delta_t, and of course it will take a longer time to find the (better) answer.
Yongjian Feng
Yongjian Feng on 16 Oct 2021
Yes, something like that. You don't need to keep the whole history. Just need to keep the previous one. So no need for y[t]. Instead, use y_total. dy_previous, dy_current....
Be carefule about ddy. Should be ddy=-g+(Cd/m)*dy_previous^2

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