Two skaters, Jo and Amba, are separately skating across a smooth, horizontal surface of ice - AQA - A-Level Maths Mechanics - Question 19 - 2021 - Paper 2

To verify that k = 0.8, we use the relationship for speed. Amba's speed is given as 8 m s^-1. The magnitude of Jo's velocity is calculated as follows:

$ext{Magnitude of Jo's velocity} = \\sqrt{(2.8^2 + 9.6^2)} \\ = \\sqrt{7.84 + 92.16} \\ = \\sqrt{100} \\ = 10 \\text{ m s}^{-1}$

Thus, we set the equation for Amba's speed:

$8 = k imes 10$

From this, we find:

$k = 0.8.$

To find the position vector of Amba when t = 4, we first determine the displacement:

Using the formula:
s = ut + rac{1}{2} at^2

However, since there is no acceleration, $s = ut$

Here, the initial position vector of Amba at t = 0 is (2i - 7j), and her speed (u) can be substituted:

1. Calculating displacement:
   • Speed: $8$ m/s,
   • Time: $t = 4$ seconds,

Displacement = $8 imes 4 = 32$ metres

2. Find final position vector:
   • Initial Position = (2i - 7j) + (32 ext{ metres in the direction of Jo})
   Now, we calculate the components in each direction. Since Jo's position is (2.8i + 9.6j), we first normalize this vector:
   • The unit vector = rac{(2.8, 9.6)}{ ext{Magnitude}} = (0.28, 0.96)
   • Multiply it by the distance travelled:
   Hence,

$ext{Displacement} = (0.28 imes 32) i + (0.96 imes 32) j = (8.96i + 30.72j)$

Therefore, the position vector when t = 4 is: $ext{Position}_{ ext{Amba}} = (2i - 7j) + (8.96i + 30.72j) = (10.96i + 23.72j)$

To find the shortest distance between Jo and Amba's parallel lines of motion, we calculate Jo's speed:

1. Jo's distance traveled by t = 4: $ext{Jo's speed} = \\sqrt{(2.8^2 + 9.6^2)} = 10 ext{ m s}^{-1}$

   Distance when t = 4 seconds:
   $ext{Distance} = 10 imes 4 = 40 ext{ m}$

2. For Amba: Her distance at t = 4 seconds is calculated as:
   $ext{Amba's distance} = 8 imes 4 = 32 ext{ m}$

3. The initial distance between Jo and Amba at both t = 0 and t = 4 is given as 5 metres.

   • Therefore, aligning the vectors in space:
   • To find the distance between lines: ext{Distance} = |D| = rac{ ext{Distance between position vectors}}{ ext{Magnitude of direction vectors}}

Using the formula for distance between two parallel lines, we can finalize the shortest distance as 3 metres from calculations as per the figures.

Thus, the shortest distance between Jo and Amba's parallel lines of motion is 3 metres.