The speed, v m s⁻¹, of a train at time t seconds is given by
v = \\sqrt{(1.2t - 1)}, \\ 0 \leq t \leq 30 - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 2
Question 8
The speed, v m s⁻¹, of a train at time t seconds is given by
v = \\sqrt{(1.2t - 1)}, \\ 0 \leq t \leq 30.
The following table shows the speed of the train at 5 sec... show full transcript
Worked Solution & Example Answer:The speed, v m s⁻¹, of a train at time t seconds is given by
v = \\sqrt{(1.2t - 1)}, \\ 0 \leq t \leq 30 - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 2
Step 1
Complete the table, giving the values of v to 2 decimal places.
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Answer
To complete the table, we need to calculate the values of v for t = 15, 20, and 25 seconds using the given formula:
For t = 15:
v=sqrt(1.2×15−1)=sqrt(18−1)=sqrt17approx4.12.
For t = 20:
v=sqrt(1.2×20−1)=sqrt(24−1)=sqrt23approx4.79.
For t = 25:
v=sqrt(1.2×25−1)=sqrt(30−1)=sqrt29approx5.39.
Use the trapezium rule, with all the values from your table, to estimate the value of s.
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Answer
To estimate the distance s travelled by the train, we use the trapezium rule with the values calculated earlier:
The formula for the trapezium rule is:
s≈2h(f(x0)+2f(x1)+2f(x2)+2f(x3)+2f(x4)+f(xn))
where h is the interval (5 seconds) and f(x) are the corresponding speed values.
Here, h = 5, the values from the table are:
f(0) = 0
f(5) = 1.22
f(10) = 2.28
f(15) = 4.12
f(20) = 4.79
f(25) = 5.39
f(30) = 6.11
Applying the trapezium rule:
s≈25(0+2×1.22+2×2.28+2×4.12+2×4.79+6.11)=25(0+2.44+4.56+8.24+9.58+6.11)=25(30.93)≈77.325.
Thus, we can estimate that the train travels approximately 154.075 meters.