observer is standing at distance d = 60 m south from the intersection
cyclist is travelling at speed v = 10 m/s
now after t = 8 s its displacement from intersection is given by
[tex]x = 10*8 = 80 m[/tex]
so the position of cyclist makes an angle with the observer
[tex]\theta = tan^{-1}\frac{80}{60} = 53 degree[/tex]
now the component of velocity of cyclist along the line joining its position with the observer is given as
[tex]v = v_o cos\phi[/tex]
here
[tex]\phi = 90 -\theta[/tex]
[tex]\phi = 90 - 53 = 37 degree[/tex]
[tex]v = 10 cos37 = 8 m/s[/tex]
so at this instant cyclist is moving away with speed 8 m/s
The cyclist moving away from the observer 8 seconds after he passes through the intersection at the rate 8 m/s
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Vector is quantity that has magnitude and direction.
One example of a vector is displacement.
Let us now tackle the problem !
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This problem is about Vector Diagram.
Given:
distance between observer and intersection = y = 60 m
speed of cyclist = v = 10 m/s
elapsed time = t = 8 s
Asked:
speed of cyclist relative to observer = v' = ?
Solution:
Firstly , we will calculate the distance between the cyclist and intersection:
[tex]x = v t[/tex]
[tex]x = 10 \times 8[/tex]
[tex]x = 80 \texttt{ m}[/tex]
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Next , we could calculate the speed of cyclist relative to observer:
[tex]v' = v \times \cos \theta[/tex]
[tex]v' = v \times \frac{ x }{\sqrt{x ^2 + y^2}}[/tex]
[tex]v' = 10 \times \frac { 80 }{\sqrt{80^2 + 60^2}}[/tex]
[tex]v' = 10 \times \frac {4}{5}[/tex]
[tex]v' = 8 \texttt{ m/s}[/tex]
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The cyclist moving away from the observer 8 seconds after he passes through the intersection at the rate 8 m/s
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Grade: High School
Subject: Physics
Chapter: Vectors
