Calculating Centripetal Acceleration in a Speed Skater's Turn

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Dylan6866
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Homework Statement



A short track speed skaer is making a turn at one end an ice rink, increasing his speed frm uniformly 10 m/s t 20 m/s in two seconds. The radius of the turn is 5 meters. While he is in the middle of the turn, and mving at 15 m/s, what is

a. tangential acceleration
b. centripetal acceleration
c. magnitude of his acceleration

Homework Equations


v^2/r
magnitude of a = sqrt((tangential)^2+(centripetal)^2)

The Attempt at a Solution



tangential= 15 m/s
centripetal=(15m/s)^2/(5m)=45m/s^2
magnitude of acceleration = sqrt((15)^2+(45)^2)=47.43

Please help me out here.
 
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Dylan6866 said:

Homework Statement



A short track speed skaer is making a turn at one end an ice rink, increasing his speed frm uniformly 10 m/s t 20 m/s in two seconds. The radius of the turn is 5 meters. While he is in the middle of the turn, and mving at 15 m/s, what is

a. tangential acceleration
b. centripetal acceleration
c. magnitude of his acceleration

Homework Equations


v^2/r
magnitude of a = sqrt((tangential)^2+(centripetal)^2)

The Attempt at a Solution



tangential= 15 m/s
centripetal=(15m/s)^2/(5m)=45m/s^2
magnitude of acceleration = sqrt((15)^2+(45)^2)=47.43

Please help me out here.
Hello Dylan6866. Welcome to PF.

If you mean that the tangential acceleration = 15m/s, I'm afraid that's wrong. Wrong units and wrong number.

The centripetal acceleration is correct.
 
SammyS said:
Hello Dylan6866. Welcome to PF.

If you mean that the tangential acceleration = 15m/s, I'm afraid that's wrong. Wrong units and wrong number.

The centripetal acceleration is correct.

Thank you. I figured it out.