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General Physics 2070 - Kinematics

by: Daria

General Physics 2070 - Kinematics 2070

Marketplace > Clemson University > Physics > 2070 > General Physics 2070 Kinematics
GPA 4.0

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About this Document

Notes covering kinematics including position, distance, and displacement. Free fall, acceleration, velocity.
General Physics 1
Amy Pope
Class Notes
kinematics, Physics, General Physics, acceleration, displacement/speed/velocity/acceleration/projective motion, velocity, average and instantaneous velocity
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This 4 page Class Notes was uploaded by Daria on Tuesday August 30, 2016. The Class Notes belongs to 2070 at Clemson University taught by Amy Pope in Fall 2016. Since its upload, it has received 6 views. For similar materials see General Physics 1 in Physics at Clemson University.


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Date Created: 08/30/16
Kinematics    Position, Distance, and Displacement  ● Before describing motion  ○ Set up a coordinate system, define an origin and positive direction, define  position  ● Position: where?  ● Distance: total of how far you go  ● Displacement: change in position = Δp  ○ Displacement to the right is positive, displacement to the left is negative      ○ Δp = 2  ○ Distance = 10    Average Speed and Velocity  ● Average speed =  elapsed time displacement ● Average velocity =  elapsed time ○ Average velocity is a vector  ○ [r] = m     [t] = s     [v] = m/s  ● Example:  ○ If a ball rolls from 10m to 0m in 2.0s, then average velocity is v= 025 = ­2.5m/s  | | ○ So what is the speed?  v | avg|= |− 2.5m/s|= 2.5m/s    Instantaneous Velocity  ● “Snapshot velocity” ­ tangent to the curve at a specific point  Δx ● v= lΔt→0 Δt    Acceleration  Δv v(final) − v(initial) ● a =  Δt= t(final) − t(initial) 2 ● [x] = m     [v] = m/s     [a] = m/s    ● Acceleration is the slope of velocity­vs­time graph       a  avg= slope =  Δt      Motion with Constant Acceleration  ● Average velocity    v avg= 2(v 0+ v)  1 2 ● Position as a function of time ­     x = 0  +0  t +2 a t    ● Velocity as a function of position ­     v   =0v  + 2 a Δt  ● Velocity as a function of time ­      v =0v   + a t    Free Fall  ● Free fall is the motion of an object in the direction of gravity.  The acceleration due to  gravity on earth is a constant, g=9.81m/s  2 ● Gravity varies slightly with altitude  ○ Gravity at the north pole ­ 9.832m/s   2 ○ Gravity on the equator ­ 9.780 m/s   ● As long as air resistance is neglected in most problems  ● Air resistance makes a big difference    Free Fall Example  ● Positive is downward direction  ○ x, v, and t are increasing  ○ Acceleration is a constant  Time  Velocity  Position  Acceleration  2 0s  v=0  x=0  x=9.81m/s    2 1s  v=9.81m/s  x=4.91m  x=9.81m/s    2s  v=19.6m/s  x=19.6m  x=9.81m/s   2 2 3s  v=29.4m/s  x=44.1m  x=9.81m/s    4s  v=39.2m/s  x=78.5m  x=9.81m/s   2   2 ● x = x 0+ v  0 +  2 t      Projectile  ● Trajectory of a projectile    Kinematic Equation Examples  ● A sports car claims that it does “zero to sixty (27m/s) in 6.2s”. How far has it traveled by  then?  ○ x = x  + v  t +  a t   2 0 0 2 ○ a =  Δt=  26.2s= 4.4m/s   2 1 2 ○ x= 0 + 0 +  (2.4) (6.2)    ○ x=83.7m  ● A package is dropped from an airplane moving upward at 15.0m/s. If it takes 16.0s  before the package strikes the ground, how high above the ground was the package  when it was released?  ○ x = x  + v  t +  a t    0 0 2 ■ x = 0  ■ x 0= ?  ■ v = Doesn’t matter for this equation  ■ v 0= 15.0m/s  2 ■ a = ­9.81m/s    ■ t = 16.0s  1 2 ○ 0 = x 0+ (15)(16) +  2­9.81) (16)    ○ x  = 1020m/s  0 ● A rock is dropped from 40m  during the first 4.0s after its release on another planet.  What is the acceleration on this planet?  ○ x = x  + v  t +  a t    0 0 2 ■ x = 40m  ■ x 0= 0  ■ v = Doesn’t matter for this equation  ■ v 0= 0m/s  ■ a = ?  ■ t = 4.0s  1 2 ○ 40 = 0 + 0 +  2 4    ○ a = 5.0m/s     


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