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The Integral and Kinematics

The Integral and Kinematics. AP Physics C Mrs. Coyle . For Constant Velocity:. Velocity (m/s). v. o. D t. Time (s). Area Under Line = Displacement D x = v D t. For a Varying Velocity. v (m/s). t(s). For a very small D t , each displacement D x = v avg D t

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The Integral and Kinematics

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  1. The Integral and Kinematics AP Physics CMrs. Coyle

  2. For Constant Velocity: Velocity (m/s) v o Dt Time (s) Area Under Line = Displacement Dx= v Dt

  3. For a Varying Velocity v (m/s) t(s) For a very small Dt, each displacement Dx = vavgDt is the small area under curve over that small Dt.

  4. The total area under the curve from ti to tf is equal to the sum of all the rectangles from ti to tf. It is written as: • vxn = instantaneous velocity, because the Dt0 • vx (t)= velocity as a function of time Limit or Integral

  5. When the limits of integration are known the integral is called a Definite Integral

  6. Integral is also known as antiderivative. • Velocity is the derivative of displacement with respect to time. v = dx dt • Displacement is the integral (antiderivative) of velocity as a function of time. dx=v dt ∆x= ∫dx = ∫v dt xf tf xi ti

  7. The definite integral of a polynomial: xf xf ∫xndx = xn+1 | n+1 Where n ≠ -1 xi xi

  8. Example 1 • For the following velocity expressions find the displacement from 2sec to 5sec. (v is in m/s, t in sec). • a) v(t)= 4t • b) v(t)= 3 • c) v(t)= t2 + 3 • d) v(t)=t-3 +1 • Ans: a) 42m, b) 9m, c) 48m, d) 3.105m

  9. Integral is also known as antiderivative. • Acceleration is the derivative of velocity with respect to time. • Change in Velocity is the integral (antiderivative) of acceleration as a function of time.

  10. Example 2: The velocity of an object is given by the relationship v= 3t2 +4. • Find the expression for the acceleration in terms of time. • Find the expression for the displacement in terms of time. • Find the displacement between 2s and 4s. • Answer for d) 64m

  11. Example 3: • The acceleration of an object is given by the function a= 6t3 –t2. • Find the expression for velocity as a function of time. Find the change in velocity between 0.2s and 2s. • Answer: 21.3 m/s

  12. Indefinite integral: • n ≠ -1 • c is a constant

  13. Example 4 –Evaluating the integral constant The acceleration of a particle along the x axis is a(t)=7.0t, (t is in sec and a is in m/s2). At t=3s, the velocity is +20m/s. What is the velocity at t=5s? Answer: 76m/s

  14. Prob. # 56 Separating variables in order to integrate. • The acceleration of a marble in a certain fluid is given by a=-3.00v2 for v>0. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble’s speed is reduced to half of its initial value? • Strategy: a=dv/dt -3.00v2 =dv/dt Separate the variables on each side: -3 dt= v-2dt , now integrate and then solve.Ans:0.222s

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