Linear Kinematics - displacement, velocity and acceleration

Contents:•Quandary•New formula•Deriving the rest of the formulas•How to solve these: suvat

• Whiteboards

it averages 20 m/s for 10 seconds, so it goes 200 m

Quandary

TOC

A car goes from 14 m/s to 26 m/s in 10. seconds. How far does it go in this time?

A new formula:

s = (u + v)t 2

TOC

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

s - displacement (m)u - initial velocity (m/s)v - final velocity (m/s)a - acceleration (m/s/s)t - time (s)

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

Deriving: - 2 more nifty formulas!!

v = u + at (memorize)

s = (u + v)t 2

From the first formula:t = (v - u)

a

Derive:v2 = u2 + 2ass = ut + 1/2at2

when to use which one…

TOC

s - displacement (m)u - initial velocity (m/s)v - final velocity (m/s)a - acceleration (m/s/s)t - time (s)

Show suvatuse v2 = u2 + 2as, and s = (u+v)t/2

Example #1

TOC

A car goes from 14 m/s to 26 m/s in 300. m. •What is the acceleration, and •What time does it take?

suvats = ut + 1/2at2 is hardv2 = u2 + 2as can be done(then use v = u + at)

Example #2

TOC

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?

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

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 + 1/2at2

s =75 m, 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 = ??

v2 = u2 + 2as

a =-9.9211 m/s/s, -9.9 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 = ??

v2 = u2 + 2as

u =29.9783255 m/s, 30. 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 + 1/2at2

(Oh Nooooooo!)OR - use v2 = u2 + 2as - find v, and use v = u + at

t =10.7585 s, 11 s

11 s W

Is itBigger than

Your Mouth?

Start

Eat it

Done

No

Yes Cut it up a bit

Basic Problem Solving

Are you done

yet?

Start

Yes

No Apply any formula

Basic Problem Solving

Write down given quantities

Circle the answer

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

2.50x103 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

21 s W