1. Physics

2. 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: v 2 = a t + v 1 aΔt = Δv v 2 – v 1 = a * Δt v 2 = aΔt + v 1

3. PHS 5042-2 Kinematics & Momentum Kinematic Equations Example: Determine the velocity of an airplane moving with constant acceleration of 5 m/s 2 along a rectilinear path after 5 minutes, if its initial velocity is 20 m/s. v 2 = at + v 1 v 2 = (5m/s 2 )(300s) + 20m/s v 2 = 1520m/s

4. PHS 5042-2 Kinematics & Momentum Kinematic Equations 2 nd Kinematic Equation: Find the variation of displacement between the interval of time highlighted on the graph. Δ d = A rectangle + A triangle Δ d = (v 1 -0)(t 2 -t 1 ) + (v 2 -v 1 )(t 2 -t 1 ) / 2 Δ d = v 1 Δ t + Δ v( Δ t ) / 2 Δ d = 10.5 m 2

5. PHS 5042-2 Kinematics & Momentum Kinematic Equations 2 nd Kinematic Equation: Δd = v 1 Δt + ΔvΔt/2 In many cases t 1 = 0, so d 2 = at 2 2 /2 + v 1 t 2 + d 1 or just: d 2 = at 2 /2 + v 1 t + d 1 Δv = aΔt Δd = v 1 Δt + aΔt Δt/2 d 2 = aΔt 2 /2 + v 1 Δt + d 1 d 2 - d 1 = v 1 Δt + aΔt 2 /2

6. PHS 5042-2 Kinematics & Momentum Kinematic 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/s 2 ? NEVER!

7. PHS 5042-2 Kinematics & Momentum Kinematic Equations 3 rd Kinematic Equation: v 2 2 - v 1 2 = 2a Δ d Example: A car moving at 50km/h needs 31m to come to a stop. What is the acceleration of the car as it breaks? v 2 2 - v 1 2 = 2aΔd a = v 2 2 - v 1 2 / 2Δd a = {[0] 2 - [(50 * 1000/3600)m/s] 2 } / 2(31m) a = - (13.9 m/s) 2 / 62m a = - 3.12 m/s 2

8. PHS 5042-2 Kinematics & Momentum Kinematic Equations Rectilinear motion with uniform acceleration Uniform rectilinear motion Main characteristica = constantv = constant Position vs time graph Position equationd 2 = at 2 /2 + v 1 t + d 1 d 2 = v 1 t + d 1 (a = 0) Velocity vs time graph Velocity equationv 2 = a t + v 1 v 2 = v 1 (a = 0) Acceleration vs time graph Other equation v 2 2 - v 1 2 = 2a Δ d (a constant) v 2 2 - v 1 2 = 0 (a = 0) P t P t P t P t a t V t V t a t V t

9. PHS 5042-2 Kinematics & Momentum Kinematic 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/s 2 )  Instead of “d”, we use “y” (Cartesian)

10. PHS 5042-2 Kinematics & Momentum Kinematic 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)

11. PHS 5042-2 Kinematics & Momentum Kinematic Equations Free fall y 2 = gt 2 /2 + v 1 t + y 1 v 2 = g t + v 1 v 2 2 - v 1 2 = 2gΔy

12. PHS 5042-2 Kinematics & Momentum Kinematic 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? v 2 2 - v 1 2 = 2gΔy v 2 2 = 2gΔy + v 1 2 v 2 2 = 2(-9.8 m/s 2 )(-0.7m) + 0 v 2 2 = √(13.72 m 2 /s 2 ) v 2 = ± 3.7 m/s v 2 = - 3.7 m/s (falling) v 2 = gΔt + v 1 Δt = v 2 - v 1 / g = v 2 / g Δt = v 2 / g Δt = (- 3.7 m/s / -9.8 m/s 2 ) Δt = 0.38 s