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A man on a motorcycle traveling at a uniform speed of 10 m/s throws an empty can straight upward relative to himself with an initial speed of 3.0 m/s. Find the equation of the trajectory as seen by a police officer on the side of the road. Assume the initial position of the can is the point where it is thrown. Ignore air resistance.

a) y = -3x + 5x²
b) y = -5x + 3x²
c) y = -3x - 5x²
d) y = 5x - 3x²

User Patentfox
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1 Answer

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Final answer:

The equation of the trajectory as seen by a police officer on the side of the road is y = -3x -4.9 * (x/10)^2.

Step-by-step explanation:

To find the equation of the trajectory as seen by a police officer on the side of the road, we need to consider the motion of the can relative to the police officer. Since the motorcycle and the can are both moving at a uniform speed of 10 m/s, the horizontal distance covered by the can is the same as the horizontal distance covered by the motorcycle. Let's call this distance 'x'.

Now, let's consider the vertical motion of the can relative to the police officer. The initial vertical velocity of the can is 3.0 m/s upwards. The only force acting on the can is gravity, so it will undergo free fall motion. The equation for the vertical motion can be written as:

y = vo * t + 0.5 * a * t^2

where y is the vertical distance, vo is the initial vertical velocity, t is the time, and a is the acceleration due to gravity. Since the can is thrown upwards, the acceleration due to gravity will be negative (-9.8 m/s^2).

Substituting the values, we get:

y = 3.0 * t - 4.9 * t^2

Now, we need to express 'y' in terms of 'x', using the fact that the horizontal distance and time are the same for the can and the motorcycle:

y = -3x -4.9 * (x/10)^2

This is the equation of the trajectory as seen by the police officer on the side of the road.

User Levan
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