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anonymous
 3 years ago
Use of numerical integral method to calculate the length of the path has traversed the car during the time recording below
anonymous
 3 years ago
Use of numerical integral method to calculate the length of the path has traversed the car during the time recording below

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Don't you have more informations concerning the method of interpolation of the velocities ? Is this linear interpolation ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, assuming linear interpolation of velocities, you can write: \[V(t) = V0 + (V1V0) t\]\[D(t) = \int\limits\limits_{0}^{t}v(x).x.dx\] Here, V0 and V1 are 2 consecutive values in your array. Like 124 and 134 for the first 2 values. D(t) is the distance you went through during time t. For this particular example of yours, we're going to always use t=6 because your time steps are constant and always equal 6 seconds. I used http://integrals.wolfram.com/index.jsp?expr=a+x+%2B+b+x%5E2&random=false to perform the integral (although it's very easy to write it yourself) so we find that: \[D(t) = \int\limits\limits_{0}^{t}v(x).x.dx\ = \left[ \frac{ V0 }{ 2 } t^2 + \frac{ V1V0 }{ 18 } t ^{3} \right]\] When solving for D(6) (t varying in [0,6]) we get: \[D(6) = 18 V0 + 12(V1V0) = 12 V1 + 6 V0\] All you need to do next is to write a loop that sums \[12 V1 + 6 V0\] for all you time intervals. It goes like: float SumDistance = 0; for ( int TimeStepIndex=0; TimeStepIndex < TimeValuesCount1; TimeStepIndex++ ) { float TimeStepDistance = 6 * Velocities[TimeStepIndex] + 12 * Velocities[TimeStepIndex+1]; SumDistance += TimeStepDistance; }

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry, my first equation should read: \[V(t) = V0 + \frac{ V1  V0 }{ 6 } t\] because starting from V0 at t=0 second, we only reach V1 after 6 seconds...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If the actual assignment was to perform numerical integration using many small time steps then I suppose my solution is wrong, instead you should go all code and write something like: int IntegrationStepsCount = 1000; // The more you use, the more precise it will get, but never as precise as an analytical integration obviously... float SumDistance = 0; for ( int TimeStepIndex=0; TimeStepIndex < TimeValuesCount1; TimeStepIndex++ ) { float TimeStepDistance = 0; float StartVelocity = Velocities[TimeStepIndex]; float EndVelocity = Velocities[TimeStepIndex+1]; float IntegrationStepSize = 6.0 / IntegrationStepsCount; for ( int i=0; i < IntegrationStepsCount ; i++ ) { float CurrentVelocity = StartVelocity + (EndVelocity  StartVelocity) * i / IntegrationStepsCount; TimeStepDistance += CurrentVelocity * IntegrationStepSize; // Distance += Velocity * DeltaTime } SumDistance += TimeStepDistance; } This is a purely numerical integration...
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