camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Usin
1 answer:
The question lacks details. Here is the full question.
The accompanying image was captured with a camera capable of shooting between one and two frames per second. A series of photos was merged into this single image, meaning the vehicles depicted are actually the same car, documented at different intervals.
Assuming the camera produced 1.3 frames per second for this image and that the length of the car is approximately 5.3 meters, based on this information and the photo, how fast was the car moving?
Answer: v = 6.5 m/s
Explanation: The problem requires calculating the car's velocity. Velocity can be computed using:
Since the camera captured 7 images of the car and its length is noted as 5.3, the car's displacement is:
Δx = 7(5.3)
Δx = 37.1 m
The camera operates at 1.3 frames per second and recorded 7 images, thus the time driven by the car is:
1.3 frames = 1 s
7 frames = Δt
Δt = 5.4 s
<pconsequently the="" car="" was="" driving="" at:="">v = 6.87 m/s
<pthe car="" moved="" at="" an="" estimated="">velocity of 6.87 m/s.
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1)
2)
Explanation:
Projectile Motion
When an object is projected near the surface of the Earth at an angle to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.
The height of a projectile can be calculated using
where represents the initial height from ground level,
is the vertical component of the initial velocity, and t is the elapsed time.
The vertical speed component is expressed as
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The peak height is attained when , which allows us to compute the time to reach that height.
Solving for
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Continuing with the calculations for
Substituting known values yields
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