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NOTES 7 - Topic 2 - Mechanics *

NOTES 7 - Topic 2 - Mechanics * - ------------------------------------------------------------------------------- 2 .1.5 Solve problems involving uniformly accelerated motion Ex ample Problem 1: An auto, starting from rest, accelerates to

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NOTES 7 - Topic 2 - Mechanics *

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  1. NOTES 7 - Topic 2 - Mechanics* -------------------------------------------------------------------------------- 2.1.5 Solve problems involving uniformly accelerated motion Example Problem 1: An auto, starting from rest, accelerates to 100. kmh-1 South in 5.0 s. (A)What is the auto’s acceleration in ms-2?(show solution in NB) A. Given: Unknown: Equation: x= xo= v= 100kmh-1= 27.8m/s vo=0 a=? t=5.0s a=? v=vo+at 27.8m/s = 0 + a (5) a=5.56 ms-2 or 6 ms-2

  2. (B) What is the auto’s displacement during this acceleration in meters? x= ? xo=0 v= 100kmh-1= 27.8m/s vo=0 a=5.56ms-2 t=5.0s Given: Unknown: Equation: x= ? x= xo + vo t + 1/2 a t2 x= 0 + 0 + 1/2 (5.56) (5.0)2 x=69.5 m ~ 70 m

  3. Example Problem 2: The best rebounders in pro basketball have a vertical leap of about 120. cm. (A) What is their initial “launch speed” from the floor? (show solution in NB) A. Given: Unknown: Equation: x= 120 cm = 1.20 m xo=0 v= 0 vo=? a=-9.8ms-2 t= vo=? v2 = v02 + 2a Δx 0=vo2 + 2 (-9.8) (1.20) vo2 = 23.52 vo = 4.85 m/s

  4. (B) How long are they in the air? x= 120 cm = 1.20 m xo=0 v= 0 vo=4.849742261m/s a=-9.8ms-2 t= Unknown: Equation: t=? v=vo+at 0=4.849742261 + (-9.8) t t=0.4948716593 s BUT this is only the time to the top which is only 1/2 the total time. So total time is 0.9897433186s ~ 1 s

  5. Example Problem 3: A stone is thrown vertically upward with a speed of 20.0 ms-1. (A) How fast is it moving when it reaches a height of 12.0 m? (show solution in NB) A. Given: Unknown: Equation: x= 12.0 m xo=0 v= vo=20m/s a=-9.8ms-2 t= v=? v2 = v02 + 2a Δx v2 = (20)2 + 2 (-9.8) (12) v2 = 164.8 v = 12.8 m/s ~ 13 m/s

  6. (B) How long is required to reach this height? x= 12.0 m xo=0 v= 12.83744523 m/s vo=20m/s a=-9.8ms-2 t= t=? Unknown: Equation: v = vo + at 12.83744 = 20 + -9.8 (t) t= 0.73087 ~ 0.73 s

  7. Example Problem 4: A person in a burning building jumps out of a 4th floor window into a fireman’s net 17.0 m below. The net is stretched by the impact by 1.50 m before stopping the person’s fall. (A) At what speed does the jumper hit the net (assume it is held 2.00 m above the ground)? (show solutions in NB) A. Given: Unknown: Equation: x= -17.0 m xo=0 v= ? vo=0 a= -9.8ms-2 t= v= ? v2 = v02 + 2a Δx v2 = 0 + 2 (-9.8) (-17) v2 = 333.2 v = ±18.3 m/s The answer should be -18.3 m/s because the final velocity is downwards.

  8. (B) What was the average acceleration experienced by the person when in contact with the net? Given: Unknown: Equation: x= -1.50 m xo=0 v= 0 vo= -18.3 m/s a=? t= a=? v2 = v02 + 2a Δx 0=(-18.3)2 + 2 (a) (-1.5m) -333.3 = -3 a a= 111.1 m/s2 ~ 111 m/s2

  9. 2.1.6 Describe the effects of air resistance on falling objects and the concept of terminal velocity • Because of air resistance, objects do not fall with the same acceleration through our atmosphere. The actual acceleration is determined by mass and surface area. Most objects falling through the atmosphere will reach a maximum velocity and stop accelerating...this maximum is called terminal velocity. At terminal velocity, the force of friction exerted by air resistance is equal to the weight of the falling object. The terminal velocity for a falling person is around 290 kmh-1... (80. ms-1= 180 miles per hour).

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