MCAT Study Guide Physics Ch. 2 – Kinematics 2017-08-15T06:45:05+00:00

I.          2.1 UNITS AND DIMENSIONS

QUANTITY SYMBOL UNITS DIMENSIONS
Speed v m/s L/T
Density ρ kg/m2 M/L3
Work W kgm2/s2 ML2/T2

II.          2.2 KINEMATICS

A.     Displacement = change in position

1.     it is the net distance an object has traveled regardless of the path it took

B.     Velocity =  speed & direction

1.     This is actually average speed

a)     vav = Δx/Δt = d/Δt

b)     average speed = (total distance)/t and can be different thav vav!!

c)     the magnitude of the velocity vector is the speed (which is scalar)

C.    Acceleration = how fast an object’s velocity changes

1.     aav = Δv/Δt

2.     an object can be accelerating even if speed is constant if it is changing direction

 

III.          2.3 UNIFORMLY ACCELERATED MOTION

A.     Big 5 equations:

  1. d = ½(v0 + v)t   missing a
  2.  v = v0 + at          missing d
  3. d = v0t + ½at2   missing v
  4. d = vt – ½at2      missing  v0
  5. v2 = v02 + 2ad   missing t

 

 IV.          2.4 KINEMATICS WITH GRAPHS

A.     Position vs time

1.     slope of this curve gives velocity

B.     Velocity vs time

1.     slope of this curve gives acceleration (antiderivative)

2.     the area under the velocity curve gives displacement (derivative)

3.     A = 2*3 = 6

4.     B = ½(3*3) = 4.5

V.          2.5 FREE FALL

A.     This involved vertical motion only

B.     a = gravitational acceleration (g) in freefall questions = 10 m/s2

 

VI.          PROJECTILE MOTION

A.     These questions involve vertical and horizontal motion – break it up into these respective parts

B.     You can figure out the velocity component of each by knowing the launch angle,  θ0

  1. Horizontal velocity = vox
    • v0x = v0cosθ
  2. Vertical velocity = v0y
    • v0y = v0sinθ
  3. Horizontal motion:
    • velocity will always be constant: v0x = v
  4. Vertical motion:
    • a = g = -10m/s2
    • v0y at the top of a parabola is zero
    • parabola is symmetrical (if it takes 3 sec to reach top, it will take 6 sec total to reach bottom)

  1. EX: cannonball is shot from ground level with initial velocity of 100 m/s at an angle of 30º
  • How high will the cannonball go?
    1. First, figure out vertical time component:
      • v0y = v0sin30º, vy = 0 (at the top of the pathway)
      • vy = v0y + at; t = (vy – v0y)/a; t = (0-v0sin30º)/(-10)
      • t = (100*-0.5)/(-10) = 5s
    2. Now, plug t = 5 into the equation to figure out dy
      • dy = v0y + ½(-g)t2
      • dy = v0sin30ºt + ½(-10)(52)
      • dy = 100(0.5)(5) -5*25 = 250 – 125 = 125 m
  • What is the velocity at the top of the path?
    1. We know vertical velocity is 0
    2. We know horizontal velocity is constant
      • v0x = v0cosθ = 100cos(30º) = 100*(0.85) = 85m/s
  • What will the total flight time be?
    1. If ½ the flight time is 5 sec, total flight time is 10 sec
  • How far will it travel horizontally?
    1. We know horizontal velocity is 85 m/s and total travel time is 10 sec
      • dy = v*t = 85*10 = 850 m

 

CHAPTER 2 SUMMARY

  • Displacement → d = △x
  • Average velocity → vav = △x/△ t = d/△t
  • Average acceleration → ā = △v/△t
  • BIG 5 EQUATIONS:
    • d = ½(v0 + v)/t
    • d = v0t + ½at2
    • d = vt – ½at2
    • v = v0 + at
    • v2 = v02 + 2ad
  • Position vs time graph → slope = velocity
  • Velocity vs time graph → slope = acceleration
    • Area under graph = displacement
  • Projectile motion:

 

Horizontal motion Vertical motion
displacement x = v0xt y = v0yt + ½(-g)t2
velocity v0x = vx → constant! vy = v0y – gt
acceleration ax = 0 ay = -g
acceleration (v0x = v0cosθ0) (voy = v0sinθy)

MCAT Study Guide Physics - Kim Matsumoto


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