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 ## VI.          PROJECTILE MOTION

### 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) # 10.

#### Ch. 11 Reflection + Refraction

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