You are watching: What is her centripetal acceleration during the turn at each end of the track?
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This is College Physics Answers with Shaun Dychko. This runner starts the race here and goes 200 meters until below, and also it takes 23.2 secs to execute that. The curved percentage of the track is this portion below and also has actually a radius of curvature of 30 meters. We"re told to number out what is the centripetal acceleration while the runner is on the curved percent. So we need to know what the runner"s linear speed is and also so we"ll take the full distance of 200 meters and also divide it by the total time to gain that. We"ll substitute that into our centripetal acceleration formula. We have actually that right here. So we have actually centripetal acceleration is v squared over r and also v in this instance is the total distance separated by the total time. So I have substituted that in for v and it gets squared and then it"s a little bit messy to have fractions within fractions and also so rather of writing every one of this over r which I can have, however rather I desired to save everything on one line so we do not have actually a fraction that"s nested within one more. So I"m multiplying by one over r instead of dividing by r. It"s much easier to view I think that you have d squared on the peak and you have actually t squared on the bottom and r on the bottom too. So this is the centripetal acceleration. The full distance, 200 meters, squared separated by the radius of curvature of 30 meters, times 23.2 secs squared, provides us 2.5 meters per second squared is the centripetal acceleration.
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