The speed, v in 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 second intervals - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 2

Question 7

$The-speed,-v-in-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-second-intervals-Edexcel-A-Level Maths Pure-Question 7-2006-Paper 2.png$

The speed, v in 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... show full transcript

Worked Solution & Example Answer:The speed, v in 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 second intervals - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 2

Step 1

Complete the table, giving the values of v to 2 decimal places.

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Answer

To find the values of v at t = 15, 25, and 30 seconds, we will use the equation:
v = \sqrt{(1.2t-1)}.

For t = 15:
$v = \sqrt{(1.2 \times 15 - 1)} = \sqrt{(18 - 1)} = \sqrt{17} \approx 4.12.$
For t = 25:
$v = \sqrt{(1.2 \times 25 - 1)} = \sqrt{(30 - 1)} = \sqrt{29} \approx 5.38.$
For t = 30:
$v = \sqrt{(1.2 \times 30 - 1)} = \sqrt{(36 - 1)} = \sqrt{35} \approx 5.92.$

Thus, the completed table is:

t	0	5	10	15	20	25	30
v	0	1.22	2.28	4.12	5.38	6.11	5.92

Step 2

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 in 30 seconds, we will apply the trapezium rule:

Using the values from our completed table:
$s \approx \frac{h}{2} \left( f(a) + 2f(x_1) + 2f(x_2) + 2f(x_3) + 2f(x_4) + 2f(x_5) + f(b) \right)$
where
h = 5,
a = 0,
b = 30,
[ f(0) = 0,\ f(5) = 1.22,\ f(10) = 2.28,\ f(15) = 4.12,\ f(20) = 5.38,\ f(25) = 6.11,\ f(30) = 5.92. ]

The values are:
$s \approx \frac{5}{2} \left( 0 + 2(1.22) + 2(2.28) + 2(4.12) + 2(5.38) + 2(6.11) + 5.92 \right)$

$\frac{5}{2} \left( 0 + 2.44 + 4.56 + 8.24 + 10.76 + 12.22 + 5.92 \right)$

$\frac{5}{2} \left( 44.14 \right) = 110.35.$
Therefore, the estimated value of s is approximately 110.35 meters.

The speed, v in 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 second intervals - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 2

Worked Solution & Example Answer:The speed, v in 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 second intervals - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 2

Complete the table, giving the values of v to 2 decimal places.

Use the trapezium rule, with all the values from your table, to estimate the value of s.

The values are: s≈52(0+2(1.22)+2(2.28)+2(4.12)+2(5.38)+2(6.11)+5.92)s \approx \frac{5}{2} \left( 0 + 2(1.22) + 2(2.28) + 2(4.12) + 2(5.38) + 2(6.11) + 5.92 \right)s≈25​(0+2(1.22)+2(2.28)+2(4.12)+2(5.38)+2(6.11)+5.92)

52(0+2.44+4.56+8.24+10.76+12.22+5.92)\frac{5}{2} \left( 0 + 2.44 + 4.56 + 8.24 + 10.76 + 12.22 + 5.92 \right)25​(0+2.44+4.56+8.24+10.76+12.22+5.92)

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The values are:
$s \approx \frac{5}{2} \left( 0 + 2(1.22) + 2(2.28) + 2(4.12) + 2(5.38) + 2(6.11) + 5.92 \right)$

$\frac{5}{2} \left( 0 + 2.44 + 4.56 + 8.24 + 10.76 + 12.22 + 5.92 \right)$