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A biker rides on an upward-sloped road inclined at a 15.0° angle with the horizontal. If he moves at a constant speed of 2.0 meters/second for 60 seconds, what is the magnitude of his horizontal displacement?

Sagot :

To determine the horizontal displacement of the biker, we'll break down the problem into several steps:

1. Identify the speed of the biker and the duration of travel:
- Speed of the biker: [tex]\( 2.0 \, \text{meters/second} \)[/tex]
- Time of travel: [tex]\( 60 \, \text{seconds} \)[/tex]

2. Understand the inclination and identify the angle:
- Angle of the road with the horizontal: [tex]\( 15.0^\circ \)[/tex]

3. Calculate the horizontal component of the biker's speed:
The biker's speed is at an angle to the horizontal, so we need the horizontal component of his speed. This component can be found using the cosine function:
[tex]\[ \text{Horizontal speed} = \text{Speed} \times \cos(\theta) \][/tex]
where [tex]\( \theta \)[/tex] is the angle of inclination ([tex]\( 15.0^\circ \)[/tex]).

Thus, we get:
[tex]\[ \text{Horizontal speed} \approx 2.0 \times \cos(15.0^\circ) \approx 1.9318516525781366 \, \text{meters/second} \][/tex]

4. Calculate the total horizontal displacement:
The horizontal displacement can then be found by multiplying the horizontal speed by the time of travel:
[tex]\[ \text{Horizontal displacement} = \text{Horizontal speed} \times \text{Time} \][/tex]
Substituting the values, we get:
[tex]\[ \text{Horizontal displacement} \approx 1.9318516525781366 \, \text{meters/second} \times 60 \, \text{seconds} \approx 115.9110991546882 \, \text{meters} \][/tex]

Thus, the horizontal displacement of the biker is approximately [tex]\(115.911 \, \text{meters}\)[/tex].