Pls help i’m not good at this lol
2 answers:
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Refer to the diagram shown below which illustrates the problem.
Initially, the runner is at A, and the observer is at B so that the distance between them is 220 m.
The angular velocity of the runner is 7 m/s)/(110 m) = 0.0636 rad/s
In 1 second, the runner sweeps an arc of 0.0636 radians. This creates a central angle of 0.0636 rad.
From the isosceles triangle created, calculate the length of AC as follows:
x = 110 sin(0.0318) = 3.4974 m
Length of AC = 2*3.4974 = 6.995 m
ΔBOC is an isosceles triangle with a vertex angle of
π - 0.0636 = 3.078 rad.
Half the length of BC is
110 sin(3.078/2) = 109.944 m
Length of BC = 2*109.944 = 219.89 m
The change in distance per second at a separation of 220m is
220 - 219.89 = 0.11 m/s
Answer: 0.11 m/s
Initially, the runner is at A, and the observer is at B so that the distance between them is 220 m.
The angular velocity of the runner is 7 m/s)/(110 m) = 0.0636 rad/s
In 1 second, the runner sweeps an arc of 0.0636 radians. This creates a central angle of 0.0636 rad.
From the isosceles triangle created, calculate the length of AC as follows:
x = 110 sin(0.0318) = 3.4974 m
Length of AC = 2*3.4974 = 6.995 m
ΔBOC is an isosceles triangle with a vertex angle of
π - 0.0636 = 3.078 rad.
Half the length of BC is
110 sin(3.078/2) = 109.944 m
Length of BC = 2*109.944 = 219.89 m
The change in distance per second at a separation of 220m is
220 - 219.89 = 0.11 m/s
Answer: 0.11 m/s