Linear Kinematics displacement velocity and acceleration Contents Quandary

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 u - initial velocity v - final velocity a - acceleration t - time (m) (m/s) (m/s/s) (s) 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 (memorize) s = (u + v)t 2 From the first formula: t = (v - u) a s - displacement u - initial velocity v - final velocity a - acceleration t - time (m) (m/s) (m/s/s) (s) Derive: v 2 = u 2 + 2 as s = ut + 1/2 at 2 when to use which one… 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 v 2 = u 2 + 2 as, 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 + 1/2 at 2 is hard v 2 = u 2 + 2 as 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. 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

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 + 1/2 at 2 s =75 m, 75 m W

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 = ? ? v 2 = u 2 + 2 as a =-9. 9211 m/s/s, -9. 9 m/s/s W

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 = ? ? v 2 = u 2 + 2 as u =29. 9783255 m/s, 30. m/s W

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 + 1/2 at 2 (Oh Nooooooo!) OR - use v 2 = u 2 + 2 as - find v, and use v = u + at t =10. 7585 s, 11 s W

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

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

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. 50 x 103 m W

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