Question:
The velocity of a particle starting from rest is given in the table. The total distance travelled in 10 s, using Trapezoidal rule, is
t (s) & 0 & 2 & 4 & 6 & 8 & 10
v (m/s) & 0 & 12 & 16 & 20 & 35 & 60
\endtabular
The velocity of a particle starting from rest is given in the table. The total distance travelled in 10 s, using Trapezoidal rule, is
t (s) & 0 & 2 & 4 & 6 & 8 & 10
v (m/s) & 0 & 12 & 16 & 20 & 35 & 60
\endtabular
v (m/s) & 0 & 12 & 16 & 20 & 35 & 60
\endtabular
Show Hint
Trapezoidal Rule = (Interval / 2) $\times$ [Sum of end terms + 2 $\times$ Sum of middle terms].
Updated On: Apr 10, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Formula Setup
Interval $h = \frac{10-0}{5} = 2$. Trapezoidal Rule: $\text{Distance} = \frac{h}{2} [v_{0} + 2(v_{1} + v_{2} + v_{3} + v_{4}) + v_{5}]$.
Step 2: Calculation
Distance $= \frac{2}{2} [0 + 2(12 + 16 + 20 + 35) + 60]$. $= [0 + 2(83) + 60] = 166 + 60 = 226$.
Final Answer: (b)
Interval $h = \frac{10-0}{5} = 2$. Trapezoidal Rule: $\text{Distance} = \frac{h}{2} [v_{0} + 2(v_{1} + v_{2} + v_{3} + v_{4}) + v_{5}]$.
Step 2: Calculation
Distance $= \frac{2}{2} [0 + 2(12 + 16 + 20 + 35) + 60]$. $= [0 + 2(83) + 60] = 166 + 60 = 226$.
Final Answer: (b)
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