Distance & Acceleration

Acceleration: Rate of change of velocity Measures how an objects velocity (or speed) is changing over time a = Change in velocity Change in time a = Δv Δt Units: m/s/s = m/s 2

Distance & Acceleration v t Δv Δt acceleration is the slope of v – t graph a = Δv Δt (+) a = increase in speed (-) a = decrease in speed {What does a = 0 mean?} (Constant Velocity!!)

Distance & Acceleration Ex. #1 A car increases its speed from 2.3 m/s to 15.6 m/s over a time 14.3 s. What is its average acceleration? Δv a = Δt a = 15.6 – a = 0.93 m/s 2

Distance & Acceleration Ex. #2 A bike slows down from 16 m/s to 1.3 m/s while accelerating at –1.2 m/s 2. How long does this acceleration take? a Δv Δt = 1.3 – Δt = 12 s (Re-arrange the original equation)

Distance & Acceleration Ex. #3 An F1 racing car can accelerate at –12.8 m/s 2 while slowing down for a curve. The car’s initial speed is 78 m/s and accelerates for 2.3 s. What is the final speed of the F1 car? Re-arrange the equation (although it looks like a new one it really isn’t!! v 2 = v 1 + at v 2 = 78 + (-12.8 x 2.3) v 2 = 49 m/s Substitution

Distance & Acceleration You need to practice the concepts that have been presented to you. 1.Read p. 63 – 69 in your text book; take the time to go through the examples given. 2.Do the following Practice problems: 1.#1 – 4 p #5 – 7 p #9 – 12 p You need to understand the above concepts in order to understand the next section. So really put some effort into this.

Distance & Acceleration Δd = v x Δt Δd = 13 x 5.0 = 65 m Case 1: Uniform Motion: What distance does a car go if it travels at a constant speed of 13 m/s for a time of 5.0 s? v = 13 m/s Δt = 5.0 s ΔdΔd

Distance & Acceleration Case 2: Non-Uniform Motion: How far does the car travel while increasing its speed from 13 m/s to 21 m/s in 5.0 s? v 2 = 21 m/s v 1 = 13 m/s Δt = 5.0 s ΔdΔd In this case a new equation is needed. Δd = v 1 + v 2 2 X Δt Where: is the average speed. v 1 + v 2 2 = v ave

Distance & Acceleration Δd = v 1 + v 2 2 X ΔtΔd = X 5.0 Δd = 17 X 5.0 Δd = 85 m {Which is 20 m more than in case 1.} New equation. substitution The math.

Distance & Acceleration Case 3: Non-Uniform Motion: How far does the car travel, starting from rest, and accelerating at 4.5 m/s 2 for 5.0 s? v 1 = 0 m/s Δt = 5.0 s Δd = ? In this case a new equation is needed. Δd = ½ at 2 Δd = ½ x 4.5 m/s 2 x (5.0 s) 2 Δd = 56 m

Distance & Acceleration Case 4: Non-Uniform Motion: How far does a car go in 5.0 s if the car has an initial velocity of 13 m/s and an acceleration of +1.9 m/s 2 ? In this case another new equation is needed. Δ d = v 1 x t + ½ x a x t 2 Note: A positive acceleration will increase distance while a negative acceleration will decrease distance. v = 13 m/s a = 1.9 m/s 2 Δt = 5.0 s Δd = ?

Distance & Acceleration Δd = Δd = 89 m New equation. substitution The math. Δ d = v 1 x t + ½ x a x t 2 Δ d = 13 x ½ x 1.9 x 5.0 2

Distance & Acceleration Case 4: Non-Uniform Motion: How far does the car now go in while accelerating at 2.6 m/s 2 while increasing its speed from 13 m/s to 35 m/s? Since there is no time in this problem another new equation is needed. v 1 = 13 m/s a = 2.6 m/s 2 v 2 = 35 m/s ΔdΔd Δ d = v 2 2 – v 1 2 2a Note: On your formula sheet this formula is in a different version: {v 2 2 = v ad}

Distance & Acceleration Δ d = v 2 2 – v 1 2 2a Δ d = 35 2 – x 2.6 Δ d = (1225 – 169) 5.2 New equation. substitution The math.

Distance & Acceleration Δ d = Δ d = 203 m Note: Use this equation when you are not given the time in the problem. More math.

Distance & Acceleration Go on to the Practice Worksheet.