Open Loop Response of a car using a Basic First Order Model

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SUMMARY

The discussion focuses on modeling the open loop response of a car using a basic first-order differential equation. The parameters include a mass (M) of 1,000 kg and a drag coefficient (D) of 1,000 kg/s. The differential equation derived is y'(t) + (D/M)y(t) = (1/M)f(t), where f(t) is a step function representing the applied force. The problem also explores the effects of varying the drag coefficient on the car's velocity over time.

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Homework Statement


Car Dynamics

f(t)→ \frac{\frac{1}{M}}{s+\frac{D}{M}}→y(t)
Applied Force Velocity

Homework Equations


M=1,000 kg and D=1000 kg/s
Where f(t) represents the input force and y(t) is the output velocity. M is the Mass and D is the drag, both of which are assumed constant for each case to be considered.

The Attempt at a Solution


The first order Differential equation is y'(t)+\frac{D}{M}y(t)=\frac{1}{M}f(t)

After I got that I'm supposed to do: I get stuck below at b part. I appreciate any help.

b) Solve the differential equation for y(t) if the input is a step function scaled by the force F0, f(t)=F0u(t). The initial velocity y(0)=28.8m/s (75 km/hr). Choose F0 such that the final velocity is 100 km/hr (27.8 m/s).

c) Plot the velocity y(t) versus time. Your time axis should go from 0 to 100 sec. Label axes.

d) How does the velocity change if the drag, D, is reduced to 75 kg/s? Increased to 150 kg/s? Plot y(t) for both cases and compare to part b above.
 
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This problem has just been posed by elijah78 and addressed by the savants of PF.
 

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