A speed skater moving across frictionless ice at 8.4 m/s hits a 5.0 m wide patch

Question

A speed skater moving across frictionless ice at 8.4 m/s hits a 5.0 m wide patch of rough ice. She slows steadily, then continues on (when she gets back on smooth ice) at 5.0 m/s. What is her acceleration on the rough ice? m/s2

HINTS: Read this question carefully. Remember you are only interested in what happens to the skater when she is slowing down on the rough patch of ice. What are the initial and final velocities for her motion on the rough patch? What do you expect the sign of the acceleration to be? The answer is NOT 4.556 m/s^2

Explanation / Answer

on the rough patch initil velocity, vi = 8.4 m/s

final velocity, vf = 5 m/s

distance travelled, d = 5 m

let a is the acceleration of skater.

use kinematic equation

vf^2 - vi^2 = 2*a*d

==> a = (vf^2 - vi^2)/(2*d)

= (5^2 - 8.4^2)/(2*5)

= -4.556 m/s^2 <<<<<----------Answer

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