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The velocity $v$ (in kilometer/minute) of a motorbike which starts form rest, is given at fixed intervals of time $t$ (in minutes) as follows:

 t 2 4 6 8 10 12 14 16 18 20 v 10 18 25 29 32 20 11 5 2 0

The approximate distance (in kilometers) rounded to two places of decimals covered in 20 minutes using Simpson's $1/3^{rd}$ rule is ________.

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is this in gate2017 syllabus?
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Nope, it is not.
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Is this in GATE 2020 syllabus ?

Topics like Newton Raphson's , Trapezoidal , Simpson Rule etc. need to be studied or not ?

Can anyone help ?

The total distance covered is given by the area under the speed curve and can be calculated by integrating the speed curve equation in the given range. In the question, exact speed function is not given, but the speed at different time instants is given.

Question is asking the distance covered in 20 minutes. According to the question, the motion starts at t = 0 and ends at t = 20. So the speed function has to be integrated in the range t=0 to t=20. Therefore an additional point (t=0) has to be considered while integrating. After adding t=0, number of points will be 11 and number of intervals will be 10. And the answer will be 309.33.

Distance

$S = \frac{(b-a)}{3n}\left [ f(0) + 4 \left( f(2) + f(6) + f(10) + f(14) + f(18) \right) \\+ 2\left ( f(4) + f(8) + f(12) + f(16) + f(20) \right)\right]$

$= \frac{2}{3} \left[0 + 4\left(10 + 25 + 32 + 11 + 2\right) \\+ 2\left(18 + 29 + 20 + 5 \right)\right]$

$= 309.33$

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How can I solve it with trapezoidal rule please, and what if it's required to solve it with a mix of the two rules
+1 vote
simply apply simpson 1/3 rule we will get 292..
but can we apply simpsons 1/3 rule when n is odd ?

edited
+1 vote