linear kinematics - acceleration contents: definition of acceleration acceleration example |...

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Linear Kinematics - Acceleration Contents: Definition of acceleration Acceleration Example | Whiteboard Example | Whiteboard

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Linear Kinematics - Acceleration

Contents:• Definition of acceleration• Acceleration

• Example | Whiteboard• Example | Whiteboard

Physics Quantities:x - Displacement - Distance and directionv - Velocity - Speed and directiont - Elapsed time

0 m-4 m +4 m

In our world, there are two directions + and -

- +x = +5 m

x = -6 m

Show velocity + and -Meter sticks/car/volvo at intersection

Whiteboards: + or - Velocity

1 | 2 | 3

TOC

A car has a velocity of -2.4 m/s for 36 seconds. What is its displacement in this time, and if it started at a position of +127.0 m, where will it end up?

v = x/tx = (-2.4 m/s)(36 s) = -86.4 m (-86 m)So we are at 127.0 + -86.4 = 40.6 m (41 m)

displacement -86 m, end up at 41 m W

A train starts at a position of -35.1 m and ends at a position of 12.5 m in a time of 13.8 seconds. What was its velocity during this time?

v = x/tx = 12.5 - -35.1 = +47.6 mv = (47.6 m)/(13.8 s) = 3.449275362 m/s 3.45 m/s

+3.45 m/s W

A train has a velocity of -25.2 m/s and ends up at a position of -57 m after a time interval of 15 seconds. What was its displacement during this time, and where did it start?

v = x/tx = x = (-25.2 m/s)(15 s) = -378 m (-380 m)Since it moved -378 m, it must have started + 378 m from -57 m, so it started at 321 m. (+320 m)

-380 m, +320 m W

Linear Kinematics -

TOC

v = x/t - time rate change of position (m/s)

a = v/t - time rate change of velocity (m/s/s)

v - Change in velocity (in m/s)a - acceleration (in m/s/s)

Example: A car goes from 0 to 27 m/s in 9.0 seconds, what is its acceleration?(show math, units, meaning, m/s/s, m/s2, ms-2)

a = v/t, a = (27 m/s)/(9.0 s) = 3.0 m/s/s

Example: a = v/tA rocket accelerates at 4.5 “g” s. What time will it take to reach the speed of sound (Mach I = 343 m/s) from rest?

1 “g” = 9.81 m/s/sso 4.5 “g”s = (4.5)(9.81 m/s/s) = 44.145 m/s/sa = v/t, 44.145 m/s/s = (343 m/s)/tt = 7.7698…t = 7.8 s

TOC

Whiteboards: Acceleration1 | 2 | 3 | 4 | 5

TOC

A car speeds up from 0 to 21 m/s in 5.3 seconds. What is their acceleration?

a = v/ta = (21 m/s)/(5.3 s) = 3.962264151 m/s/s4.0 m/s/s

4.0 m/s/s W

A train can accelerate at .15 m/s/s. What time will it take to reach its top speed of 24 m/s from rest?

160 s W

a = v/tv = 24 m/s, a = .15 m/s/s, .:. t = 160 s160 s

What is the final speed if a person accelerates from rest at 32 f/s/s for 2.7 seconds?

86 f/s W

a = v/tt = 2.7 s, a = 32 f/s/s, .:. v = 86.4 f/s86 f/s

What is your acceleration if your velocity goes from 35 m/s to 20. m/s in 4.7 seconds?

-3.2 m/s/s W

a = v/tv = -15 m/s, t = 4.7 s, .:. a = -3.1915-3.2 m/s/s

What is your final velocity if you are going 12 m/s and you accelerate at .48 m/s/s for the next 16 seconds?

20. m/s W

a = v/ta = .48 m/s/s, t = 16 s, .:. v = 7.68 m/s12 m/s + 7.68 m/s = 19.68 m/s 20. m/s

Consider the last problem:12 m/s + 7.68 m/s = 19.68 m/s vi + at = vf

New formula:vf = vi + at

vi - initial velocityvf - final velocitya - accelerationt = time elapsedShow + and - acceleration with + and – velocitycars have a + and – accelerator pedalwhat happens when you get that wrong…..

TOC

Whiteboards: vf = vi + at

1 | 2 | 3 | 4 | 5

TOC

A car going 24.8 m/s decelerates at -2.451 m/s/s for 1.67 s. What is its final velocity?

vf = vi + atvi = 24.8 m/s, a = -2.451 m/s/s, t = 1.67 s .:. vf = 20.707 m/s

20.7 m/s

20.7 m/s W

What is the acceleration of a ball that goes from 12.5 m/s to 17.8 m/s in 2.5 seconds?

vf = vi + atvi = 12.5 m/s, vf = 17.8 m/s, t = 2.5 s .:. a = 2.12 m/s/s 2.1 m/s/s

2.1 m/s/s W

A cop clocks a car going 22.4 m/s after having accelerated at -7.45 m/s/s for the last 3.4 seconds. What was the initial velocity of the car?

vf = vi + atvf = 22.4 m/s, a = -7.45 m/s/s, t = 3.4 s .:. vi = 47.73 m/s

48 m/s

48 m/s W

What time will it take a train to slow from 23.2 m/s to 14.8 m/s if the acceleration is -1.2 m/s/s?

vf = vi + atvi = 23.2 m/s, vf = 14.8 m/s, a = -1.2 m/s/s, .:. t = 7 s 7.0 s

7.0 s W

A pitcher throws a ball at a velocity at +34.2 m/s. The batter does a line drive, giving the ball a velocity of -45.1 m/s. If the ball was in contact with the bat for .2153 seconds, what was its acceleration during that time?

vf = vi + atvi = +34.2 m/s, vf = -45.1 m/s, t = .2153 s, .:. a = -368.32327 m/s/s -368 m/s/s

-368 m/s/s W