Search Menu

Contents

Problems for Position, Velocity, and Acceleration as Vectors

Problems for Position, Velocity, and Acceleration as Vectors

Problems for Position, Velocity, and Acceleration as Vectors

Problems for Position, Velocity, and Acceleration as Vectors

Problems for Position, Velocity, and Acceleration as Vectors

Problems for Position, Velocity, and Acceleration as Vectors

Problem : Find the derivative of the vector-valued function,

f(x) = (3x 2 +2x + 23, 2x 3 +4x, x -5 +2x 2 + 12)

We take the derivative of a vector-valued function coordinate by coordinate:

f'(x) = (6x + 2, 6x 2 +4, -5x -4 + 4x)

Problem : The motion of a creature in three dimensions can be described by the following equations for position in the x -, y -, and z -directions.


x(t) = 3t 2 + 5  
y(t) = - t 2 + 3t - 2  
z(t) = 2t + 1  

Find the magnitudes** of the acceleration, velocity, and position vectors at times t = 0 , t = 2 , and t = - 2 .

The first order of business is to write the above equations in vector form. Because they are all (at most quadratic) polynomials in t , we can write them together as:

x(t) = (3, -1, 0)t 2 + (0, 3, 2)t + (5, - 2, 1)

We are now in a position to compute the velocity and acceleration functions. Using the rules established in this section we find that,


v(t) = 2(3, - 1, 0)t + (0, 3, 2) = (6, - 2, 0)t + (0, 3, 2)  
a(t) = (6, - 2, 0)  

Notice that the acceleration function a(t) is constant; therefore the magnitude (and direction!) of the acceleration vector will be the same at all times:

|a| = |(6, -2, 0)| = = 2

All that's left now is to compute the magnitudes of the position and velocity vectors at times t = 0, 2, - 2 :
  • At t = 0 , |x(0)| = |(5, -2, 1)| = , and |v(0)| = |(0, 3, 2)| =
  • At t = 2 , |x(2)| = |(17, 0, 5)| = , and |v(2)| = |(12, -1, 2)| =
  • At t = - 2 , |x(- 2)| = |(17, -12, -3)| = , and |v(- 2)| = |(- 12, 7, 2)| =
Notice that the magnitude of the creature's velocity (i.e. the speed at which the creature is traveling) is high at t = - 2 , decreases considerably at t = 0 , and goes back up again at t = 2 , even though the acceleration is constant! This is because the acceleration is causing the creature to slow down and change direction--in the same way that a ball thrown upwards (which experiences constant acceleration due to the earth's gravity) slows down to zero-velocity as it reaches its maximum height, and then changes direction to fall back down.