a car travels at constant speed aroudn a circular trach whose radius is 3.2 km the car goes once around the track in 360 s what is the magnitude of the centripetal acceleration of the car?
please help me set this up so i can solve thank you sooo mcuh
2007-02-11 05:25:00 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
First convert the radius of the circle to meters. Then remember two things:
1) v=d/t = (2pi r)/T
and
2) Acentripetal = v^2 / R
You should then be able to find the car's speed, V, with the radius of the track and the time to make it around once. And then you should be able to calculate the magnitude of the centripetal acceleration.
(Note: the direction of the centripetal acceleration is toward the center of the circle.)
2007-02-11 05:34:54 · answer #1 · answered by Dennis H 4 · 0⤊ 0⤋
centripetal accel = v^2/r
v = d/t = 2pi * r / 360s
so,
v^2/r = 4*pi^2 * r /360^2
2007-02-11 13:35:16 · answer #2 · answered by animal 2 · 0⤊ 0⤋
a = (4 * pi^2 * r) / T^2
change 3.2km to m
3.2km = 3200m
a = (4 * pi^2 * 3200m) / 360s^2
a= 126,330.9 / 129,6300
a= .9748m/s^2
2007-02-11 13:34:00 · answer #3 · answered by 7 · 1⤊ 0⤋