Physics. PHS 5042-2 Kinematics & Momentum Kinematic Equations 1 st Kinematic Equation: a = Δv / Δt...

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Physics

Transcript of Physics. PHS 5042-2 Kinematics & Momentum Kinematic Equations 1 st Kinematic Equation: a = Δv / Δt...

Page 1: Physics. PHS 5042-2 Kinematics & Momentum Kinematic Equations 1 st Kinematic Equation: a = Δv / Δt In many cases t 1 = 0, so v 2 = a t 2 + v 1 or just:

Physics

Page 2: Physics. PHS 5042-2 Kinematics & Momentum Kinematic Equations 1 st Kinematic Equation: a = Δv / Δt In many cases t 1 = 0, so v 2 = a t 2 + v 1 or just:

PHS 5042-2 Kinematics & MomentumKinematic Equations

1st Kinematic Equation:

a = Δv / ΔtIn many cases t1 = 0, so

v2 = a t2 + v1

or just:v2 = a t + v1

aΔt = Δv

v2 – v1 = a * Δt

v2 = aΔt + v1

Page 3: Physics. PHS 5042-2 Kinematics & Momentum Kinematic Equations 1 st Kinematic Equation: a = Δv / Δt In many cases t 1 = 0, so v 2 = a t 2 + v 1 or just:

PHS 5042-2 Kinematics & MomentumKinematic Equations

Example:Determine the velocity of an airplane moving with constant acceleration of 5 m/s2 along a rectilinear path after 5 minutes, if its initial velocity is 20 m/s.

v2 = at + v1

v2 = (5m/s2)(300s) + 20m/sv2 = 1520m/s

Page 4: Physics. PHS 5042-2 Kinematics & Momentum Kinematic Equations 1 st Kinematic Equation: a = Δv / Δt In many cases t 1 = 0, so v 2 = a t 2 + v 1 or just:

PHS 5042-2 Kinematics & MomentumKinematic Equations

2nd Kinematic Equation:

Find the variation of displacement between the interval of time highlighted on the graph.

Δd = A rectangle + A triangle

Δd = (v1-0)(t2-t1) + (v2-v1)(t2-t1) / 2

Δd = v1Δt + Δv(Δt ) / 2

Δd = 10.5 m2

Page 5: Physics. PHS 5042-2 Kinematics & Momentum Kinematic Equations 1 st Kinematic Equation: a = Δv / Δt In many cases t 1 = 0, so v 2 = a t 2 + v 1 or just:

PHS 5042-2 Kinematics & MomentumKinematic Equations

2nd Kinematic Equation:

Δd = v1Δt + ΔvΔt/2

In many cases t1 = 0, sod2 = at2

2/2 + v1t2 + d1

or just:d2 = at2/2 + v1t + d1

Δv = aΔt

Δd = v1Δt + aΔt Δt/2

d2 = aΔt2/2 + v1Δt + d1

d2 - d1 = v1Δt + aΔt2/2

Page 6: Physics. PHS 5042-2 Kinematics & Momentum Kinematic Equations 1 st Kinematic Equation: a = Δv / Δt In many cases t 1 = 0, so v 2 = a t 2 + v 1 or just:

PHS 5042-2 Kinematics & MomentumKinematic Equations

Example:

A car moves 5m down a ramp with an initial velocity of 1m/s. How long before it reaches the bottom of the ramp if it moves with constant acceleration of 2m/s2?

d2 = 1/2at2 + v1t + d1

0 = ½ (-2m/s2)t2 + (-1m/s)t + 5m

0 = ½ (2m/s2)t2 + (1m/s)t - 5m

0 = (1m/s2)t2 + (1m/s)t - 5m

t = 1.79 s t = -2.79s

NEVER!

Page 7: Physics. PHS 5042-2 Kinematics & Momentum Kinematic Equations 1 st Kinematic Equation: a = Δv / Δt In many cases t 1 = 0, so v 2 = a t 2 + v 1 or just:

PHS 5042-2 Kinematics & MomentumKinematic Equations

3rd Kinematic Equation:v2

2 - v12 = 2aΔd

Example:A car moving at 50km/h needs 31m to come to a stop. What is the acceleration of the car as it breaks?

v22 - v1

2 = 2aΔda = v2

2 - v12 / 2Δd

a = {[0]2 - [(50 * 1000/3600)m/s]2} / 2(31m)a = - (13.9 m/s)2 / 62m

a = - 3.12 m/s2

Page 8: Physics. PHS 5042-2 Kinematics & Momentum Kinematic Equations 1 st Kinematic Equation: a = Δv / Δt In many cases t 1 = 0, so v 2 = a t 2 + v 1 or just:

PHS 5042-2 Kinematics & MomentumKinematic Equations

Rectilinear motion with uniform acceleration

Uniform rectilinear motion

Main characteristic a = constant v = constant

Position vs time graph

Position equation d2 = at2/2 + v1t + d1 d2 = v1t + d1 (a = 0)

Velocity vs time graph

Velocity equation v2 = a t + v1 v2 = v1 (a = 0)

Acceleration vs time graph

Other equation v22 - v1

2 = 2aΔd (a constant)

v22 - v1

2 = 0 (a = 0)

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Page 9: Physics. PHS 5042-2 Kinematics & Momentum Kinematic Equations 1 st Kinematic Equation: a = Δv / Δt In many cases t 1 = 0, so v 2 = a t 2 + v 1 or just:

PHS 5042-2 Kinematics & MomentumKinematic Equations

Free fall

Vertical falling movement due to the pull of the Earth Uses same motion equations than rectilinear

movement with uniform acceleration Instead of “a”, we use “g” (g = - 9.8 m/s2) Instead of “d”, we use “y” (Cartesian)

Page 10: Physics. PHS 5042-2 Kinematics & Momentum Kinematic Equations 1 st Kinematic Equation: a = Δv / Δt In many cases t 1 = 0, so v 2 = a t 2 + v 1 or just:

PHS 5042-2 Kinematics & MomentumKinematic Equations

Free fall

Upward movement: positive direction

(displacement, velocity, acceleration) Downward movement: negative direction

(displacement, velocity, acceleration) d = 0 (y = 0) ground, d > 0 (positive, above

ground), d < 0 (negative, below ground)

Page 11: Physics. PHS 5042-2 Kinematics & Momentum Kinematic Equations 1 st Kinematic Equation: a = Δv / Δt In many cases t 1 = 0, so v 2 = a t 2 + v 1 or just:

PHS 5042-2 Kinematics & MomentumKinematic Equations

Free fall

y2 = gt2/2 + v1t + y1

v2 = g t + v1

v22 - v1

2 = 2gΔy

Page 12: Physics. PHS 5042-2 Kinematics & Momentum Kinematic Equations 1 st Kinematic Equation: a = Δv / Δt In many cases t 1 = 0, so v 2 = a t 2 + v 1 or just:

PHS 5042-2 Kinematics & MomentumKinematic Equations

Example:A key falls from a height of 70 cm. What is its velocity when it hits the ground?

A key falls from a height of 70 cm. What is the duration of the flight?

v22 - v1

2 = 2gΔyv2

2 = 2gΔy + v12

v22 = 2(-9.8 m/s2)(-0.7m) + 0

v22 = √(13.72 m2/s2)

v2 = ± 3.7 m/sv2 = - 3.7 m/s (falling)

v2 = gΔt + v1 Δt = v2 - v1 / g = v2 / g Δt = v2 / g Δt = (- 3.7 m/s / -9.8 m/s2) Δt = 0.38 s