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Alenkinab [10]
3 years ago
5

Now determine the precise time that the cars started their trip (hour:minute:second) if Car 1 passed the streetlight at 12:25:00

and was 25 seconds before Car 2. Hint: Determine how far the cars traveled and then use the known velocity of Car 1 to estimate how much travel time it needed to go that distance.
Physics
1 answer:
vlada-n [284]3 years ago
6 0

This question is incomplete, the missing part is in the image below;

Answer:

the precise time that the cars started their trip is 12:00:00

Explanation:

Given that;

from the image; The velocity of car1 V1 = 1/60 mile/seconds

Given the precise time that the car 1 passed the streetlight ( 12:25:00 )

if car 1 passed 25 seconds before car 2

meaning Car 2 passed the the streetlight at 12:50:00

now time delay is the difference between the two times; i.e

12:50:00 - 12:25:00 = 25 sec

so the distance between starting point and streetlight = 25×1 = 25 miles

Now, time required for Car1 to travel 25 miles distance will be;

t = distance/velocity of car1 = 25 / (\frac{1}{60} ) = 1500 sec = 25 min

so the precise time that the cars started their trip will be;

12:25:00 - 25 min = 12:00:00

Therefore, the precise time that the cars started their trip is 12:00:00

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Answer:

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2. The magnitude of direction of the vector is 21.8°.

Explanation:

From the question given, we obtained the following data:

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Displacement along y (Bᵧ) = 10 m

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Magnitude of direction (θ) =.?

1. Determination of the magnitude of displacement of the vector (B)

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Displacement along y (Bᵧ) = 10 m

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Displacement along y (Bᵧ) = 10 m

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The magnitude of direction of the vector can be obtained by using Tan ratio as illustrated below:

Adjacent = Bₓ = 25 m

Opposite = Bᵧ = 10 m

Magnitude of direction (θ) =.?

Tan θ = Opposite /Adjacent

Tan θ = Bᵧ / Bₓ

Tan θ = 10 /25

Tan θ = 0.4

Take the inverse of Tan

θ = Tan¯¹ (0.4)

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