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What is the magnitude of the centripetal acceleration of the car?

A car travels at a constant speed around a circular track whose radius is 3.74 km. The car goes once around the track in 353 s.

a = v^2/r

vel = distance traveled/time

distance = 2 pi r = 2 pi x 3740 m = 23,487.2m

vel = 23487.2/353s=66.5m/s

a=66.5^2/3740m = 1.18m/s/s

This entry was posted on Tuesday, October 27th, 2009 at 3:05 am and is filed under 353. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “What is the magnitude of the centripetal acceleration of the car?”

  1. a = v^2/r

    vel = distance traveled/time

    distance = 2 pi r = 2 pi x 3740 m = 23,487.2m

    vel = 23487.2/353s=66.5m/s

    a=66.5^2/3740m = 1.18m/s/s
    References :

  2. Yeah, the car simply go around the track. Going the track’s just a law of nature
    References :

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