The velocity \(v\) (in kilometer/minute) of a motorbike which starts from…
2015
The velocity \(v\) (in kilometer/minute) of a motorbike which starts from 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 1/3rd rule is _______________.
Attempted by 14 students.
Show answer & explanation
Correct answer: 308 to 310
Key idea: include the velocity at t = 0 (which is 0 since the motorbike starts from rest) and apply Simpson's 1/3 rule on the interval [0, 20] with step h = 2 minutes.
Values used (t in minutes, v in km/min):
t = 0, v = 0
t = 2, v = 10; t = 4, v = 18; t = 6, v = 25; t = 8, v = 29; t = 10, v = 32
t = 12, v = 20; t = 14, v = 11; t = 16, v = 5; t = 18, v = 2; t = 20, v = 0
Computation:
Number of subintervals n = 10 (even), step h = 2.
Sum of velocities at odd indices (t = 2,6,10,14,18): 10 + 25 + 32 + 11 + 2 = 80.
Sum of velocities at even interior indices (t = 4,8,12,16): 18 + 29 + 20 + 5 = 72.
Apply Simpson's 1/3 rule: distance = (h/3)[v0 + v10 + 4(sum of odd-index velocities) + 2(sum of even-index interior velocities)].
Distance = (2/3)[0 + 0 + 4(80) + 2(72)] = (2/3)[320 + 144] = (2/3)(464) = 928/3 ≈ 309.333...
Final answer (rounded to two decimals): 309.33 km