Formulae

# Formulae - down the plane a s = ±g Sin q Example of...

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Formulae KINEMATICS Motion with constant acceleration a s : a s = D v s / D t = (v fs – v is )/ D t; Therefore , v fs = v is + a s D t. Sf = s i + (area under velocity curve v s between t i and t f ) = s i + v is Dt + ½a s (Dt) 2 s i + (v fs 2 - v is 2 )/a s or v fs 2 = v is 2 + 2a s Ds For constant acceleration

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Velocity from acceleration: Area under the v/t graph An frictionless inclined plane can be used to “decrease” the acceleration due to gravity. The component of g perpendicular to the surface (g Cos q ) is cancelled by the normal force from the plane. The component of g parallel to the surface (g Sinq) acts to accelerate the object

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Unformatted text preview: down the plane. a s = ±g Sin q Example of forumula: A skier’s speed at the bottom of a 100 m slope is 20 m/s. What is the angle of the slope? v 2s 2 = v 1s 2 +2 a s Dx = 2 g Sin q Dx, so Sin q = v 2s 2 /(2 g Dx) = (20 m/s) 2 /[2(9.80 m/s 2 )(100 m)] = 0.204 q = Sin-1 (0.204) = 11.8 O VECTORS Forces A= F/m The force is directly proportional while the mass is inversely proportional M is always positive. W=mg Where W is weight, m is mass and g is gravitational force of 9.8m/s 2...
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## This note was uploaded on 04/28/2008 for the course PHY 132 taught by Professor Rijssenbeek during the Spring '04 term at SUNY Stony Brook.

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Formulae - down the plane a s = ±g Sin q Example of...

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