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Kinematics-Average Values

Kinematics-Average Values. Average velocity = D x/ D t Average acceleration = D v/ D t. In the limit that D x goes to zero, these become instantaneous values: v = dx/dt a = dv/dt. Kinematics. a = dv/ dt v = dx/ dt

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Kinematics-Average Values

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  1. Kinematics-Average Values Average velocity = Dx/Dt Average acceleration = Dv/Dt In the limit that Dx goes to zero, these become instantaneous values: v = dx/dt a = dv/dt

  2. Kinematics a = dv/dt v = dx/dt If you know x(t) you can get v(t) and a(t) Example: Find v(t) and a(t), if x(t) = 10 +20t +5t2 V(t) = dx/dt= d/dt(10 +20t +5t2 ) = 20 + 10t a(t) = dv/dt = d/dt( 20 + 10t) = 10

  3. Example The position of a particle is given by: x(t) = 10t – 5t2 At what time(s) is the particle at x = 0? What is the velocity at t = 3 seconds? What is the average velocity for the first 4 seconds?

  4. Kinematics If you know a(t) you can get v(t) and x(t). • Example: a(t) = 5t Find v(t) and x(t). • ; = (5/2) t2 ; (5/2) t2 • ; x= ; • x+

  5. Example The acceleration vs. time graph for a particle starting from rest is given above. 1. What velocity change did the particle undergo in 10 seconds? 2. What distance did the particle move through in 10 seconds?

  6. If a = constant, v vs. t is a straight line. The average value of v is halfway on the line, so: Vavg = ½(vi +vf) = (xf – xi )/ t ,or xf = xi + (t/2)(vi +vf)

  7. So far: 1. vf = vi + at 2. xf = xi + (t/2)(vi +vf) Solving 1. for t and inserting into 2 gives you : xf = xi + (vf -vi )(vi +vf)(a/2) Or vf2 = vi 2+2a(xf –xi)

  8. So far: 1. vf = vi + at 3. vf2 = vi 2+2a(xf –xi) 2. xf = xi + (t/2)(vi +vf) Substituting 1. into 2. gives: xf = xi + t/2(vi +vi +at), or xf = xi + vit + 1/2at2

  9. Finally, the Four Kinematic Equations 1. vf = vi + at 3. vf2 = vi 2+2a(xf –xi) 2. xf = xi + (t/2)(vi +vf) 4. xf = xi + vit + (1/2)at2

  10. How To Use Them 1st List the possible unknowns xf = vf = a = xi = vi = t = 2nd Read the problem and fill in all you can ( usually 4 ) 3rd Choose a kinematic equation with just one unknown in it.

  11. Example A car starts from rest and accelerates to 40 m/s in 10 seconds. xf = ? vf = 40 a = ? xi = 0 vi = 0 t = 10 What is the acceleration of the car? How far does the car move?

  12. To get a, you need an equation with a in it, but without xf. Which one is it? 1. vf = vi + at 40 = 0 + 10a or a = 4.00 m/s2

  13. To get xfyou have a choice.Using 2., xf = xi + (t/2)(vi +vf) xf = 0 + (10/2)(0 +40) = 5(40) = 200m

  14. Summary 1. vf = vi + at 3. vf2 = vi 2+2a(xf –xi) 2. xf = xi + (t/2)(vi +vf) 4. xf = xi + vit + (1/2)at2

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