Physics, 10.05.2021 19:50 tmcdowell69
An unmarked police car traveling a constant 36.8 m/s is passed by a speeder traveling 56 m/s. Precisely 2.8 seconds after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is a constant 1.2 m/s2, how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed).
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The x-coordinate of a particle in curvilinear motion is given by x = 3.1t3 - 4.9t where x is in feet and t is in seconds. the y-component of acceleration in feet per second squared is given by ay = 2.3t. if the particle has y-components y = 0 and vy = 5.0 ft/sec when t = 0, find the magnitudes of the velocity v and acceleration a when t = 1.8 sec. sketch the path for the first 1.8 seconds of motion, and show the velocity and acceleration vectors for t = 1.8 sec. answers: v = ft/sec a = ft/sec2
Answers: 2
An unmarked police car traveling a constant 36.8 m/s is passed by a speeder traveling 56 m/s. Precis...
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