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Motion in One Dimension Notes and Example Problems

Motion in One Dimension Notes and Example Problems. D t = time in seconds D x= horizontal distance in meters D y = vertical distance in meters v i = the initial velocity in m/s v f = the final velocity in m/s a = acceleration in m/s 2. The Variables:. Acceleration of Gravity.

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Motion in One Dimension Notes and Example Problems

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  1. Motion in One DimensionNotes and Example Problems

  2. Dt = time in seconds Dx= horizontal distance in meters Dy = vertical distance in meters vi = the initial velocity in m/s vf = the final velocity in m/s a = acceleration in m/s2 The Variables:

  3. Acceleration of Gravity • If an object is falling towards the earth surface then you use g or ag = -9.81 m/s2 . *This means that an object speeds up 9.8 m/s every second. *It is negative because objects fall downward. • If the object is simply moving, use normal equations to solve for displacement and acceleration.

  4. vav= Dx Dt a = (vf -vi)Dt Dy = vi Dt + ½ aDt2Dy or Dx, depending on what direction the object is accelerating Dx = ½(vi + vf )Dt vf = vi + aDt vf2 = vi2 + 2aDx The Equations:

  5. Problem Solving Tips • Anything going down is • Object starting from rest • Object that is stopping • Object at its peak when thrown up • When dropping an object, we know • Time up = time down in vertical displacement

  6. Problem Set-Up • Given: • Dt = 0.53s not stated, but still there vi = 0 m/s • a = -9.8 m/s2 • Unknown: • Dy • Equation: • Dy = vi Dt + 1/2 aDt2

  7. Solution vi= 0 m/s so Dy = vi Dt + 1/2 aDt2 since vi = 0 Dy = 1/2 aDt2 Dy = 1/2 (-9.8 m/s2)(0.53s)2 Dy = -1.376 m (falling down) • So the table is approximately 1.4 meters high.

  8. Tom needs to get away from a bomb that Jerry has set. If he uniformly accelerates from rest at a rate of 1.5 m/s2, what will his final velocity be when he gets 25 meters away?

  9. Tom and Jerry Problem Set-Up • Given: vi = 0 m/s Dx= 25 m a = 1.5 m/s 2 • Unknown: vf • Equation: vf2 = vi2 + 2aDx

  10. Tom and Jerry Problem Solution vf2 = vi2 + 2aDx vf2 = (0 m/s)2 + 2(1.5 m/s2)(25 m) vf2 = 75 m2/s2 vf = 8.7 m/s Tom will be traveling at 8.7 m/s.

  11. Problem Set-Up • Given: vi = 16 m/s vf = 0 m/s Dx = 63 m • Unknown: a • Equation: vf2 = vi2 + 2a D x

  12. Solution A negative acceleration is the same thing as a deceleration. vf2 = vi2 + 2a D x (0 m/s)2 = (16 m/s)2 + 2 a (63 m) - 256 m2/s2 =2 a (63 m) 2 a = - 256 m2/s2 63 m 2 a = -4.06 m/s2 a = -2.03 m/s2 a = -2.0 m/s2

  13. Road Runner Problem Set-Up • Given: vi = 0 m/s a= -9.8 m/s2 Dt = 3.5 s • Unknown: vf and Dy • Equation: a = (vf - vi)/ Dt

  14. Road Runner Problem Solution PART a) a = (vf -vi)/Dt but vi = 0 m/s so (-9.8 m/s2) = (vf - 0 m/s)/ 3.5 s (-9.8 m/s2) = vf / 3.5 s = -34.3 m/s The piano will hit the ground at a speed of -34 m/s, neglecting the affects of air resistance of course.

  15. Road Runner Problem Solution PART b) • Unknown: Dy • Equation: Dy = vi Dt + 1/2 aDt2

  16. Road Runner Problem Solution PART b) Dy = vi Dt + 1/2 aDt2 but vi = 0 m/s so Dy = 1/2 a Dt2 = 1/2 (-9.8 m/s2)(3.5 s) 2 = -60.025 m The piano fell 60. meters.

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