Second Order Homogenous Differential Equation help
A car of mass 800kg is travelling along a road at a constant velocity of 100kmh^-1. A catastrophic engine and brake failure renders the car unable to brake in any way other than natural slow-down from the friction from the air and road. The friction is modelled as a backwards force equal to 40000 times the velocity (in kmh^-1).
Show that the equation of motion of the car as it slows (x in kilometres, t in hours) simplifies to d^2x/dt^2 + 50 (dx/dt) = 0.
Really not sure how to do this.
Show that the equation of motion of the car as it slows (x in kilometres, t in hours) simplifies to d^2x/dt^2 + 50 (dx/dt) = 0.
Really not sure how to do this.
Scroll to see replies
Original post by TAEuler
A car of mass 800kg is travelling along a road at a constant velocity of 100kmh^-1. A catastrophic engine and brake failure renders the car unable to brake in any way other than natural slow-down from the friction from the air and road. The friction is modelled as a backwards force equal to 40000 times the velocity (in kmh^-1).
Show that the equation of motion of the car as it slows (x in kilometres, t in hours) simplifies to d^2x/dt^2 + 50 (dx/dt) = 0.
Really not sure how to do this.
Show that the equation of motion of the car as it slows (x in kilometres, t in hours) simplifies to d^2x/dt^2 + 50 (dx/dt) = 0.
Really not sure how to do this.
use N2L
Original post by BobbJo
use N2L
What's N2L? Newton's second law?
Original post by TAEuler
What's N2L? Newton's second law?
yup
resultant force = mass x acceleration
Original post by BobbJo
yup
resultant force = mass x acceleration
resultant force = mass x acceleration
So what? It says a constant velocity, so I know you have to use the 800kg. But that'd be 800 x 0? Or is it F=400000v, hence = 40000 x 100.
hence F = 4,000,000.
Then what?
Original post by TAEuler
So what? It says a constant velocity, so I know you have to use the 800kg. But that'd be 800 x 0? Or is it F=400000v, hence = 40000 x 100.
hence F = 4,000,000.
Then what?
hence F = 4,000,000.
Then what?
no, it's moving at constant velocity before the failure
when the failure occurs, the only force on the car is the resistive force which is proportional to the velocity
resistive force is not constant
Original post by BobbJo
no, it's moving at constant velocity before the failure
when the failure occurs, the only force on the car is the resistive force which is proportional to the velocity
resistive force is not constant
when the failure occurs, the only force on the car is the resistive force which is proportional to the velocity
resistive force is not constant
But it says "a car of mass 800kg is travelling along a road at a constant velocity of 100kmh^-1"?
Right so, F is directly proportional to V? Hence F = 40000v. Then what? I'm so confused
Original post by TAEuler
But it says "a car of mass 800kg is travelling along a road at a constant velocity of 100kmh^-1"?
Right so, F is directly proportional to V? Hence F = 40000v. Then what? I'm so confused
Right so, F is directly proportional to V? Hence F = 40000v. Then what? I'm so confused
Read whole question:
"A car of mass 800kg is travelling along a road at a constant velocity of 100kmh^-1. A catastrophic engine and brake failure renders the car unable to brake in any way other than natural slow-down from the friction from the air and road. The friction is modelled as a backwards force equal to 40000 times the velocity (in kmh^-1).
Show that the equation of motion of the car as it slows (x in kilometres, t in hours) simplifies to d^2x/dt^2 + 50 (dx/dt) = 0."
Particularly, look at last sentence
Use N2L now:
F = ma
Original post by TAEuler
But it says "a car of mass 800kg is travelling along a road at a constant velocity of 100kmh^-1"?
Right so, F is directly proportional to V? Hence F = 40000v. Then what? I'm so confused
Right so, F is directly proportional to V? Hence F = 40000v. Then what? I'm so confused
F = ma. So you can find the acceleration (as a function of v).
Original post by BobbJo
Read whole question:
"A car of mass 800kg is travelling along a road at a constant velocity of 100kmh^-1. A catastrophic engine and brake failure renders the car unable to brake in any way other than natural slow-down from the friction from the air and road. The friction is modelled as a backwards force equal to 40000 times the velocity (in kmh^-1).
Show that the equation of motion of the car as it slows (x in kilometres, t in hours) simplifies to d^2x/dt^2 + 50 (dx/dt) = 0."
Particularly, look at last sentence
Use N2L now:
F = ma
"A car of mass 800kg is travelling along a road at a constant velocity of 100kmh^-1. A catastrophic engine and brake failure renders the car unable to brake in any way other than natural slow-down from the friction from the air and road. The friction is modelled as a backwards force equal to 40000 times the velocity (in kmh^-1).
Show that the equation of motion of the car as it slows (x in kilometres, t in hours) simplifies to d^2x/dt^2 + 50 (dx/dt) = 0."
Particularly, look at last sentence
Use N2L now:
F = ma
I don't really get what you mean. And I don't know how I'd construct the equation from that.
I think I want to find a, which has to be negative, but I'm just unsure of what to actually do.
Sudden brainstorm whilst writing this!
Just realised dx/dt (s) = v
and dx/dt (v) = a
But how do I construct the equation using the information I have?
Original post by TAEuler
I don't really get what you mean. And I don't know how I'd construct the equation from that.
I think I want to find a, which has to be negative, but I'm just unsure of what to actually do.
Sudden brainstorm whilst writing this!
Just realised dx/dt (s) = v
and dx/dt (v) = a
But how do I construct the equation using the information I have?
I think I want to find a, which has to be negative, but I'm just unsure of what to actually do.
Sudden brainstorm whilst writing this!
Just realised dx/dt (s) = v
and dx/dt (v) = a
But how do I construct the equation using the information I have?
yup the second derivative of x wrt t is a
They asked the equation of motion when the car slows
So once the engine failure occurs, the velocity begins to decrease due to resistive forces
ma = resistive forces
you have to pay attention to the sign, since a and v are vectors
Original post by TAEuler
I don't really get what you mean. And I don't know how I'd construct the equation from that.
I think I want to find a, which has to be negative, but I'm just unsure of what to actually do.
Sudden brainstorm whilst writing this!
Just realised dx/dt (s) = v
and dx/dt (v) = a
But how do I construct the equation using the information I have?
I think I want to find a, which has to be negative, but I'm just unsure of what to actually do.
Sudden brainstorm whilst writing this!
Just realised dx/dt (s) = v
and dx/dt (v) = a
But how do I construct the equation using the information I have?
So, F = 40000v, using F = ma, with m = 800.
40000v = -800a
Hence a = -50v
Therefore a + 50v = 0
And then use what I've said about differentiating v to get a?? I might be wrong but is a = -50v, or a = 50v? Or can I say because it's decreasing a must be negative?
Original post by TAEuler
So, F = 40000v, using F = ma, with m = 800.
40000v = -800a
Hence a = -50v
Therefore a + 50v = 0
And then use what I've said about differentiating v to get a?? I might be wrong but is a = -50v, or a = 50v? Or can I say because it's decreasing a must be negative?
40000v = -800a
Hence a = -50v
Therefore a + 50v = 0
And then use what I've said about differentiating v to get a?? I might be wrong but is a = -50v, or a = 50v? Or can I say because it's decreasing a must be negative?
Yes, a is negative since velocity is decreasing
a + 50v = 0 is the answer
just use the relation you said in your previous post to write it in the form needed
Original post by TAEuler
A car of mass 800kg is travelling along a road at a constant velocity of 100kmh^-1. A catastrophic engine and brake failure renders the car unable to brake in any way other than natural slow-down from the friction from the air and road. The friction is modelled as a backwards force equal to 40000 times the velocity (in kmh^-1).
Show that the equation of motion of the car as it slows (x in kilometres, t in hours) simplifies to d^2x/dt^2 + 50 (dx/dt) = 0.
Really not sure how to do this.
Show that the equation of motion of the car as it slows (x in kilometres, t in hours) simplifies to d^2x/dt^2 + 50 (dx/dt) = 0.
Really not sure how to do this.
May I know, what exam board this is?
Original post by TurboZX453
May I know, what exam board this is?
It's from a Hodder Education Textbook for Year 2 Further Maths. Is AQA approved and for the AQA syllabus, so would say AQA, why?
Original post by TurboZX453
May I know, what exam board this is?
CIE
Original post by TAEuler
It's from a Hodder Education Textbook for Year 2 Further Maths. Is AQA approved and for the AQA syllabus, so would say AQA, why?
I too am a Further Maths student of Edexcel and we cove the same topics in which you apply Differential Equations to modelling so I was curious. Forgive me for not helping you out as I haven’t gone over the topic yet and I am stuck too!
I need help with the following question if anyone is happy to help?:
In this question you will prove the result of the general solution of a second order differential equation for which the auxiliary equation has repeated roots.
i) find the general solution of the differential equation d^2y/dx^2 = 0:
therefore, lambda^2 = 0
so lambda = 0
I KNOW that the general solution to this is y = A + Bx
But the fact it says i'm proving this, I'm not sure if I can go straight to this to finish the question. Can I?
In this question you will prove the result of the general solution of a second order differential equation for which the auxiliary equation has repeated roots.
i) find the general solution of the differential equation d^2y/dx^2 = 0:
therefore, lambda^2 = 0
so lambda = 0
I KNOW that the general solution to this is y = A + Bx
But the fact it says i'm proving this, I'm not sure if I can go straight to this to finish the question. Can I?
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