1 / 118

VECTOR FUNCTIONS

13. VECTOR FUNCTIONS. VECTOR FUNCTIONS. 13.4 Motion in Space: Velocity and Acceleration. In this section, we will learn about: The motion of an object using tangent and normal vectors. MOTION IN SPACE: VELOCITY AND ACCELERATION.

erica-whitaker
Télécharger la présentation

VECTOR FUNCTIONS

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 13 VECTOR FUNCTIONS

  2. VECTOR FUNCTIONS 13.4 Motion in Space: Velocity and Acceleration In this section, we will learn about: The motion of an object using tangent and normal vectors.

  3. MOTION IN SPACE: VELOCITY AND ACCELERATION • Here, we show how the ideas of tangent and normal vectors and curvature can be used in physics to study: • The motion of an object, including its velocity and acceleration, along a space curve.

  4. VELOCITY AND ACCELERATION • In particular, we follow in the footsteps of Newton by using these methods to derive Kepler’s First Law of planetary motion.

  5. VELOCITY • Suppose a particle moves through space so that its position vector at time t is r(t).

  6. VELOCITY Vector 1 • Notice from the figure that, for small values of h, the vector approximates the direction of the particle moving along the curve r(t).

  7. VELOCITY • Its magnitude measures the size of the displacement vector per unit time.

  8. VELOCITY • The vector 1 gives the average velocity over a time interval of length h.

  9. VELOCITY VECTOR Equation 2 • Its limit is the velocity vector v(t) at time t :

  10. VELOCITY VECTOR • Thus, the velocity vector is also the tangent vector and points in the direction of the tangent line.

  11. SPEED • The speed of the particle at time tis the magnitude of the velocity vector, that is, |v(t)|.

  12. SPEED • This is appropriate because, from Equation 2 and from Equation 7 in Section 13.3, we have: • = rate of change of distance with respect to time

  13. ACCELERATION • As in the case of one-dimensional motion, the acceleration of the particle is defined as the derivative of the velocity: a(t) = v’(t) = r”(t)

  14. VELOCITY & ACCELERATION Example 1 • The position vector of an object moving in a plane is given by: r(t) = t3 i + t2 j • Find its velocity, speed, and acceleration when t = 1 and illustrate geometrically.

  15. VELOCITY & ACCELERATION Example 1 • The velocity and acceleration at time tare: • v(t) = r’(t) = 3t2i + 2t j • a(t) = r”(t) = 6t I +2 j

  16. VELOCITY & ACCELERATION Example 1 • The speed at t is:

  17. VELOCITY & ACCELERATION Example 1 • When t = 1, we have: v(1) = 3 i + 2 j a(1) = 6 i + 2 j|v(1)| =

  18. VELOCITY & ACCELERATION Example 1 • These velocity and acceleration vectors are shown here.

  19. VELOCITY & ACCELERATION Example 2 • Find the velocity, acceleration, and speed of a particle with position vector r(t) = ‹t2, et, tet›

  20. VELOCITY & ACCELERATION Example 2

  21. VELOCITY & ACCELERATION • The figure shows the path of the particle in Example 2 with the velocity and acceleration vectors when t = 1.

  22. VELOCITY & ACCELERATION • The vector integrals that were introduced in Section 13.2 can be used to find position vectors when velocity or acceleration vectors are known, as in the next example.

  23. VELOCITY & ACCELERATION Example 3 • A moving particle starts at an initial position r(0) = ‹1, 0, 0›with initial velocity v(0) = i – j +k • Its acceleration is a(t) = 4t i + 6t j + k • Find its velocity and position at time t.

  24. VELOCITY & ACCELERATION Example 3 • Since a(t) = v’(t), we have: • v(t) = ∫a(t) dt =∫ (4t i + 6t j + k) dt • =2t2i + 3t2j + t k + C

  25. VELOCITY & ACCELERATION Example 3 • To determine the value of the constant vector C, we use the fact that v(0) = i –j + k • The preceding equation gives v(0) = C. • So, C = i – j +k

  26. VELOCITY & ACCELERATION Example 3 • It follows: • v(t) = 2t2i + 3t2j + t k + i – j + k • = (2t2 + 1) i + (3t2 – 1) j + (t + 1) k

  27. VELOCITY & ACCELERATION Example 3 • Since v(t) = r’(t), we have: • r(t) = ∫v(t) dt • = ∫ [(2t2 + 1) i + (3t2 – 1) j + (t + 1) k] dt • = (⅔t3 + t) i + (t3 – t) j + (½t2 + t) k + D

  28. VELOCITY & ACCELERATION Example 3 • Putting t = 0, we find that D = r(0) = i. • So,the position at time t is given by: • r(t) = (⅔t3 + t + 1) i + (t3 – t)j + (½t2 + t) k

  29. VELOCITY & ACCELERATION • The expression for r(t) that we obtained in Example 3 was used to plot the path of the particle here for 0 ≤t ≤ 3.

  30. VELOCITY & ACCELERATION • In general, vector integrals allow us to recover: • Velocity, when acceleration is known • Position, when velocity is known

  31. VELOCITY & ACCELERATION • If the force that acts on a particle is known, then the acceleration can be found from Newton’s Second Law of Motion.

  32. VELOCITY & ACCELERATION • The vector version of this law states that if, at any time t, a force F(t) acts on an object of mass m producing an acceleration a(t), then • F(t) = ma(t)

  33. VELOCITY & ACCELERATION Example 4 • An object with mass m that moves in a circular path with constant angular speed ω has position vector r(t) = a cos ωt i + a sin ωt j • Find the force acting on the object and show that it is directed toward the origin.

  34. VELOCITY & ACCELERATION Example 4 • To find the force, we first need to know the acceleration: • v(t) = r’(t) = –aω sin ωt i + aω cos ωt j • a(t) = v’(t) = –aω2cos ωt i – aω2sin ωt j

  35. VELOCITY & ACCELERATION Example 4 • Therefore, Newton’s Second Law gives the force as: • F(t) = ma(t) • = –mω2 (a cos ωt i + a sin ωt j)

  36. VELOCITY & ACCELERATION Example 4 • Notice that: F(t) = –mω2r(t) • This shows that the force acts in the direction opposite to the radius vector r(t).

  37. VELOCITY & ACCELERATION Example 4 • Therefore, it points toward the origin.

  38. CENTRIPETAL FORCE Example 4 • Such a force is called a centripetal (center-seeking) force.

  39. VELOCITY & ACCELERATION Example 5 • A projectile is fired with: • Angle of elevation α • Initial velocity v0

  40. VELOCITY & ACCELERATION Example 5 • Assuming that air resistance is negligible and the only external force is due to gravity, find the position function r(t) of the projectile.

  41. VELOCITY & ACCELERATION Example 5 • What value of α maximizes the range (the horizontal distance traveled)?

  42. VELOCITY & ACCELERATION Example 5 • We set up the axes so that the projectile starts at the origin.

  43. VELOCITY & ACCELERATION Example 5 • As the force due to gravity acts downward, we have: F = ma = –mgjwhere g = |a|≈ 9.8 m/s2. • Therefore, a = –gj

  44. VELOCITY & ACCELERATION Example 5 • Since v(t) = a, we have: • v(t) = –gt j + C • where C = v(0) = v0. • Therefore, r’(t) = v(t) = –gt j + v0

  45. VELOCITY & ACCELERATION Example 5 • Integrating again, we obtain: • r(t) = –½ gt2j + t v0 + D • However, D = r(0) = 0

  46. VELOCITY & ACCELERATION E. g. 5—Equation 3 • So, the position vector of the projectile is given by: r(t) = –½gt2j + t v0

  47. VELOCITY & ACCELERATION Example 5 • If we write |v0| = v0 (the initial speed of the projectile), then • v0 = v0 cos αi + v0 sin αj • Equation 3 becomes: r(t) = (v0 cos α)t i + [(v0 sin α)t – ½gt2] j

  48. VELOCITY & ACCELERATION E. g. 5—Equations 4 • Therefore, the parametric equations of the trajectory are: x = (v0 cos α)t y = (v0 sin α)t – ½gt2

  49. VELOCITY & ACCELERATION Example 5 • If you eliminate t from Equations 4, you will see that y is a quadratic function of x.

  50. VELOCITY & ACCELERATION Example 5 • So, the path of the projectile is part of a parabola.

More Related