Two skaters, Jo and Amba, are separately skating across a smooth, horizontal surface of ice - AQA - A-Level Maths Pure - Question 19 - 2021 - Paper 2
Question 19
Two skaters, Jo and Amba, are separately skating across a smooth, horizontal surface of ice.
Both are moving in the same direction, so that their paths are straight... show full transcript
Worked Solution & Example Answer:Two skaters, Jo and Amba, are separately skating across a smooth, horizontal surface of ice - AQA - A-Level Maths Pure - Question 19 - 2021 - Paper 2
Step 1
Explain why Amba's velocity must be in the form $k(2.8i + 9.6j)$ m s$^{-1}$
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Answer
Amba's velocity must be in the form k(2.8i+9.6j) m s−1 because both skaters are moving in the same direction and their paths are parallel to each other. A vector can be expressed as a scalar multiple of another vector when they are parallel. Therefore, Amba's velocity is a scaled version of Jo's velocity, which can be written as vA=kimesvJ.
Step 2
Verify that $k = 0.8$
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To verify that k=0.8, we use the given speed of Amba, which is 8 m s−1.
Since Amba's velocity is expressed as:
extvelocityA=k(2.8i+9.6j)
We can find the magnitude of Jo's velocity:
∣vJ∣=extsqrt(2.82+9.62)=extsqrt(7.84+92.16)=extsqrt(100)=10extms−1
Now, we set up the equation:
∣vA∣=kimes∣vJ∣8=kimes10
Thus, k=0.8.
Step 3
Find the position vector of Amba when $t = 4$
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To find Amba's position vector at t=4, we first calculate the displacement using the formula:
s=ut
Where u is Amba's velocity:
vA=0.8(2.8i+9.6j)=(2.24i+7.68j)
Now, the displacement after 4 seconds is:
s=(2.24i+7.68j)imes4=(8.96i+30.72j)
Since Amba's initial position at t=0 is (2i−7j), we add the displacement:
rA=(2i−7j)+(8.96i+30.72j)=(2+8.96)i+(−7+30.72)j=(10.96i+23.72j)
So, Amba's position vector when t=4 is (10.96i+23.72j).
Step 4
Determine the shortest distance between Jo and Amba's positions
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Answer
We first find Jo's position at t=4:
extJo′sposition=(2.8i+9.6j)imes4=(11.2i+38.4j)
Next, we calculate the distance d between Jo and Amba:
The vector from Jo to Amba at t=4 is:
extVector=extJo′sposition−extAmba′sposition=(11.2i+38.4j)−(10.96i+23.72j)=(11.2−10.96)i+(38.4−23.72)j=0.24i+14.68j
The distance d is:
d=extsqrt((0.24)2+(14.68)2)
ightarrow ext{approx. } 14.68 ext{ m}$$
Finally, to find the shortest distance between the parallel lines, we need to subtract the given distance (5 metres) from this value:
$$d_{ ext{shortest}} = 14.68 - 5 = 9.68 ext{ m}$$.
