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Accelerated Motion Problems

Accelerated Motion Problems. 1. An automobile manufacturer claims that its sports car can accelerate from rest to a speed of 39.0 m/s in 8.0 seconds. Determine the acceleration of the car a = v f –v i / Δ t = 39.0 m/s/8.0 seconds = 4.9 m/s 2

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Accelerated Motion Problems

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  1. Accelerated Motion Problems 1. An automobile manufacturer claims that its sports car can accelerate from rest to a speed of 39.0 m/s in 8.0 seconds. • Determine the acceleration of the car • a = vf –vi/Δt = 39.0 m/s/8.0 seconds = 4.9 m/s2 • b) Find the distance the car would travel in those 8 seconds • X = ½ at2 + vit = ½ (4.9m/s2)(8s)2 • = 157 m

  2. Problem #2 • A racing car starting from rest accelerates at a rate of 4.5 m/s2. What is the velocity after it has traveled for 10 seconds? Vf = at + vi = 4.5 m/s2 (10s) = 45 m/s+ 0

  3. Problem #3 • A train running at 30 m/s is slowed uniformly to a stop in 44 seconds. Find the acceleration and stopping distance. a = vf – vi/t = 0 – 30 m/s / 44 seconds = -0.68 m/s2 x = ½ at2 + vit = ½ (-.68 m/s2)(44s)2 + 30 m/s (44s) =-658 m + 1320 m = 662 m

  4. Problem #4 • A car is traveling at 60 km/h. It accelerates at 3m/s2 for 5 seconds. What speed does it reach? 60 km/h x 1000m/1km x 1h/3600 s = 16.7 m/s vf = at + vi = 3m/s2 (5s) + 16.7 m/s = 31.7 m/s

  5. Problem #5 • An object falls freely from rest. Find the distance it falls in 3.5 seconds. Find the final speed it reaches after falling this time. a = 9.8 m/s2 x = ½ at2 + vit x = ½ (9.8 m/s2)(3.5s)2 = 60 m

  6. Acceleration Problems set #2 There are four basic equations for accelerated motion problems. If you identify the given information, you can apply the appropriate equation vf = at + vi x = ½ at2 + vit x = ½ (vi + vf) 2ax = vf2 – vi2

  7. Acceleration problemsset #2 • An airplane must have a velocity of 50 m/s in order to take off. What must be the airplane’s acceleration if the runway is 500 meters long? vf = 50 m/s x= 500 m vi = 0 a=? 2ax = vf2 –vi2 a = vf2 –vi2/2x a = 2.5 m/s2

  8. Acceleration problemsset #2 • A car traveling 10 m/s accelerates to a speed of 25 m/s in 12 seconds. What distance does it cover in that time? vi = 10 m/s vf = 25 m/s t = 12 seconds x=? X = ½( vf + vi)t = ½( 10m/s + 25m/s) 12 seconds x = 210 meters

  9. Acceleration problemsset #2 • A car moving at 35 m/s is brought to a stop in 12 seconds. What is the acceleration? What distance does it cover? vf=0 vi = 35 m/s t = 12 s a = vf – vi/t = -35 m/s/12 s a = -2.9 m/s2 x = ½(vf+vi)t = 210 meters

  10. Projectile motion • Key Idea – The horizontal motion is independent of the vertical motion.

  11. Projectile Motion • Horizontally launched projectile. The vertical motion is identical to free fall

  12. Projectile Motion • Example: A motorcycle rider jumps off a cliff. His horizontal motion is described by: • X=vxt • His vertical motion is described by Y = 1/2gt2

  13. Projectile motion Example: A motorcycle rider jumps off a 35 meter high cliff at 45 m/s. How far from the edge of the cliff does he land?

  14. Projectile motion Example #2 Example: An airplane moving at a constant velocity of 115 meters/second and altitude of 1050 meters drops a package. How far horizontally does it fall from its point of release?

  15. Projectile motion Example #3 For the package released from the airplane, what is its speed just before it hits the ground?

  16. Projectile Motion • Projectiles launched at an angle: • The initial velocity must first be resolved into vertical and horizontal components Vy = vsinθ Vx = vcosθ

  17. Projectile Motion • Example#4 A placekicker kicks a football at an angle of 40 degrees with a speed of 22 m/s. Find the time of flight, range, and maximum height. Vx = vcosθ = 22m/s(cos40) = 16.9 m/s Vy = vsinθ = 22m/s(sin40) = 14.1 m/s T (time to highest point) = vy/g = 1.4 seconds Time of flight = 2T = 2.8 seconds Range = vx 2T = 16.9 m/s (2.8 seconds) = 47.3 meters Maximum height Y = 1/2gt2 + vyt =-1/2 (9.8m/s2)(1.4s)2 + 14.1m/s(1.4 s) = 10.1 meters

  18. Home Run? • A ball player hits a pitch 30m/s at 20o, does it clear a fence 5 meters high, 90 meters away? First resolve the velocity into x and y components • vx = vcos20o = 28.2 m/s • Vy = vsin20o = 10.3 m/s • The time we need to find is the time to the fence tf tf = 90m/ 28.2 m/s = 3.2 seconds How high is the ball at that time? Y =-1/2 gt2 +vyt = -50.1 m + 33 m = -17.1 m The ball never reaches the fence.

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