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

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

Physics Quantities: x - Displacement - Distance and direction v - Velocity- Speed and direction t - 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 11 | 2 | 323 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/  t  x = (-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/  t  x = 12.5 - -35.1 = +47.6 m v = (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/  t  x =  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/s 2, ms -2 ) a =  v/t, a = (27 m/s)/(9.0 s) = 3.0 m/s/s

Example: a =  v/t A 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/s so 4.5 “g”s = (4.5)(9.81 m/s/s) = 44.145 m/s/s a =  v/t, 44.145 m/s/s = (343 m/s)/t t = 7.7698… t = 7.8 s TOC

Whiteboards: Acceleration 11 | 2 | 3 | 4 | 52345 TOC

A car speeds up from 0 to 21 m/s in 5.3 seconds. What is their acceleration? a =  v/  t a = (21 m/s)/(5.3 s) = 3.962264151 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/  t  v = 24 m/s, a =.15 m/s/s,.:. t = 160 s 160 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/  t t = 2.7 s, a = 32 f/s/s,.:.  v = 86.4 f/s 86 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/  t  v = -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/  t a =.48 m/s/s, t = 16 s,.:.  v = 7.68 m/s 12 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 v i + at = v f New formula: v f = v i + at v i - initial velocity v f - final velocity a - acceleration t = time elapsed Show + and - acceleration with + and – velocity cars have a + and – accelerator pedal what happens when you get that wrong….. TOC

Whiteboards: v f = v i + at 11 | 2 | 3 | 4 | 52345 TOC

A car going 24.8 m/s decelerates at -2.451 m/s/s for 1.67 s. What is its final velocity? v f = v i + at v i = 24.8 m/s, a = -2.451 m/s/s, t = 1.67 s.:. v f = 20.707 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? v f = v i + at v i = 12.5 m/s, v f = 17.8 m/s, t = 2.5 s.:. a = 2.12 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? v f = v i + at v f = 22.4 m/s, a = -7.45 m/s/s, t = 3.4 s.:. v i = 47.73 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? v f = v i + at v i = 23.2 m/s, v f = 14.8 m/s, a = -1.2 m/s/s,.:. t = 7 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? v f = v i + at v i = +34.2 m/s, v f = -45.1 m/s, t =.2153 s,.:. a = -368.32327 m/s/s -368 m/s/s W

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