3233 Words13 Pages

VECTOR FUNCTIONS
VECTOR FUNCTIONS
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
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.
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.
VELOCITY
Suppose a particle moves through space so that its position vector at
time t is r(t).
VELOCITY
Vector 1
Notice from the figure*…show more content…*

So, C=i–j+ k VELOCITY & ACCELERATION Example 3 It follows: v(t) = 2t2 i + 3t2 j + t k + i – j + k = (2t2 + 1) i + (3t2 – 1) j + (t + 1) k 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 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 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. VELOCITY & ACCELERATION In general, vector integrals allow us to recover: Velocity, when acceleration is known v(t ) v(t0 ) t t0 a(u ) du Position, when velocity is known r (t ) r (t0 ) t t0 v(u ) du 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. 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) VELOCITY & ACCELERATION Example 4 An object with mass m that moves in a circular path with

So, C=i–j+ k VELOCITY & ACCELERATION Example 3 It follows: v(t) = 2t2 i + 3t2 j + t k + i – j + k = (2t2 + 1) i + (3t2 – 1) j + (t + 1) k 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 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 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. VELOCITY & ACCELERATION In general, vector integrals allow us to recover: Velocity, when acceleration is known v(t ) v(t0 ) t t0 a(u ) du Position, when velocity is known r (t ) r (t0 ) t t0 v(u ) du 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. 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) VELOCITY & ACCELERATION Example 4 An object with mass m that moves in a circular path with

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