Linear Kinematics - displacement, velocity and acceleration Contents: Quandary New formula Deriving the rest of the formulas How to solve these: suvat Whiteboards

Quandary A car goes from 14 m/s to 26 m/s in 10. seconds. How far does it go in this time? it averages 20 m/s for 10 seconds, so it goes 200 m TOC

A new formula: s = (u + v)t 2 s - displacement (m) u - initial velocity (m/s) v - final velocity (m/s) a - acceleration (m/s/s) t - time (s) s = (u + v)t 2 A car goes from 14 m/s to 26 m/s in 10. seconds. How far does it go in this time? s = (14 m/s + 26 m/s)(10. s) = 200 m = 2.0 x 102 m 2 A person starts from rest and goes 14.5 m in 3.75 s. What is their final velocity? 14.5 m = (0 + v)(3.75 s), v = 7.73333 = 7.73 m/s 2 TOC

Deriving: - 2 more nifty formulas!! v = u + at s = (u + v)t 2 From the first formula: t = (v - u) a Derive: v2 = u2 + 2as s = ut + 1/2at2 when to use which one… s - displacement (m) u - initial velocity (m/s) v - final velocity (m/s) a - acceleration (m/s/s) t - time (s) TOC

Example #1 A car goes from 14 m/s to 26 m/s in 300. m. What is the acceleration, and What time does it take? Show suvat use v2 = u2 + 2as, and s = (u+v)t/2 TOC

Example #2 A rocket going 3130 m/s accelerates at 0.00135 m/s/s for a distance of 5.50 x 109 m. What time does it take, and What is the final velocity? suvat s = ut + 1/2at2 is hard v2 = u2 + 2as can be done (then use v = u + at) TOC

Whiteboards: suvat 1 | 2 | 3 | 4 TOC

A cart stops in a distance of 3. 81 m in a time of 4. 51 s A cart stops in a distance of 3.81 m in a time of 4.51 s. What was its initial velocity? s = 3.81 m, u = ??, v = 0, t = 4.51 s s = (u + v)t 2 u =1.68958 m/s, 1.69 m/s W 1.69 m/s

A car going 12 m/s accelerates at 1. 2 m/s/s for 5. 0 seconds A car going 12 m/s accelerates at 1.2 m/s/s for 5.0 seconds. What is its displacement during this time? s = ?, u = 12 m/s, v = ??, a = 1.2 m/s/s, t = 5.0 s s = ut + 1/2at2 s =75 m, 75 m W 75 m

Another car with a velocity of 27 m/s stops in a distance of 36. 74 m Another car with a velocity of 27 m/s stops in a distance of 36.74 m. What was its acceleration? s = 36.74 m, u = 27 m/s, v = 0, a = ??, t = ?? v2 = u2 + 2as a =-9.9211 m/s/s, -9.9 m/s/s W -9.9 m/s/s

A car’s brakes slow it at 9. 5 m/s/s. If it stops in 47 A car’s brakes slow it at 9.5 m/s/s. If it stops in 47.3 m, how fast was it going to start with? s = 47.3 m, u = ??, v = 0, a = -9.5 m/s/s, t = ?? v2 = u2 + 2as u =29.9783255 m/s, 30. m/s W 30. m/s

Another different car has a final velocity of 12 Another different car has a final velocity of 12.0 m/s after having decelerated for a distance of 78.0 m in 4.50 seconds. What was the acceleration of the car? (Hint – use no a for u, then no s for a) solution W -2.37 m/s/s

What time will it take a car going 23 m/s to start with, and accelerating at 3.5 m/s/s, to go 450 m? s = 450 m, u = 23 m/s, v = ??, a = 3.5 m/s/s, t = ?? s = ut + 1/2at2 (Oh Nooooooo!) OR - use v2 = u2 + 2as - find v, and use v = u + at t =10.7585 s, 11 s W 11 s

Start Basic Problem Solving Is it Bigger than Your Mouth? Yes Cut it up a bit No Eat it Done

Start Basic Problem Solving Write down given quantities Are you done yet? No Apply any formula Yes Circle the answer

What distance does a train go if it slows from 32. 0 m/s to 5 What distance does a train go if it slows from 32.0 m/s to 5.0 m/s in 135 seconds? s = ?, u = 32.0 m/s, v = 5.0 m/s, t = 135 s s = (u + v)t 2 s =2497.5 m, 2.50x103 m W 2.50x103 m

A ship slows from 18 m/s to 12 m/s over a distance of 312 m A ship slows from 18 m/s to 12 m/s over a distance of 312 m. What time did it take? s = 312 m, u = 18 m/s, v = 12 m/s, t = ?? s = (u + v)t 2 t =20.8 s, 21 s W 21 s