Motion in Space

Motion in space describes a moving object with vectors: position r(t), velocity v(t) = r′(t), speed ‖v(t)‖ and acceleration a(t) = r″(t). Acceleration splits into a tangential part a_T = (v · a)/‖v‖ that changes the speed and a normal part a_N = ‖v × a‖/‖v‖ that changes the direction.

Before this lesson: Curvature, Unit Tangent and Unit Normal Vectors and Position, Velocity and Acceleration

Velocity, speed and acceleration in 3D

In Calculus 1, an object moved along a number line, and velocity and acceleration were single numbers. In space they are vectors. The velocity v(t) points the way the object moves, and its length is the speed. The acceleration a(t) shows how the velocity is changing.

Acceleration can change velocity in two ways. Pressing the gas pedal changes your speed: that part of a points along T. Turning the steering wheel changes your direction: that part points along N, toward the inside of the turn. Every acceleration is a mix of the two.

Motion in space formulas

\mathbf{v}(t) = \mathbf{r}'(t), \qquad \text{speed} = \|\mathbf{v}(t)\|, \qquad \mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t)

\mathbf{a} = a_T\,\mathbf{T} + a_N\,\mathbf{N}, \qquad a_T = \dfrac{\mathbf{v} \cdot \mathbf{a}}{\|\mathbf{v}\|} = \dfrac{d}{dt}\|\mathbf{v}\|, \qquad a_N = \dfrac{\|\mathbf{v} \times \mathbf{a}\|}{\|\mathbf{v}\|} = \kappa\,\|\mathbf{v}\|^2

• a_T is positive when the object speeds up, negative when it slows down, and 0 at constant speed.
• a_N is never negative. It is 0 on a straight path and large on a tight turn at high speed.
• The two parts satisfy ‖a‖² = a_T^2 + a_N^2, a useful check.
• Going backward, integrate a(t) to get v(t), then integrate v(t) to get r(t), using an initial condition each time.

Motion in space examples

Example 1: Velocity, speed and acceleration

An object moves along r(t) = ⟨2t, t², t³/3⟩. Find v(t), the speed, and a(t), and evaluate them at t = 1.

\mathbf{v}(t) = \langle 2, \; 2t, \; t^2 \rangle

\|\mathbf{v}(t)\| = \sqrt{4 + 4t^2 + t^4}

= \sqrt{(t^2 + 2)^2} = t^2 + 2

\mathbf{a}(t) = \langle 0, \; 2, \; 2t \rangle

\mathbf{v}(1) = \langle 2, 2, 1 \rangle, \quad \|\mathbf{v}(1)\| = 3, \quad \mathbf{a}(1) = \langle 0, 2, 2 \rangle

Answer: v(t) = ⟨2, 2t, t²⟩, speed t² + 2, and a(t) = ⟨0, 2, 2t⟩. At t = 1 the velocity is ⟨2, 2, 1⟩, the speed is 3, and the acceleration is ⟨0, 2, 2⟩.

Example 2: Tangential and normal components

For the object in Example 1, find a_T and a_N at t = 1, and check that a_T^2 + a_N^2 = ‖a‖².

\mathbf{v} \cdot \mathbf{a} = (2)(0) + (2)(2) + (1)(2)

= 0 + 4 + 2 = 6

a_T = \dfrac{6}{3} = 2

\mathbf{v} \times \mathbf{a} = \langle 2(2) - 1(2), \; 1(0) - 2(2), \; 2(2) - 2(0) \rangle

= \langle 2, \; -4, \; 4 \rangle

\|\mathbf{v} \times \mathbf{a}\| = \sqrt{4 + 16 + 16} = 6

a_N = \dfrac{6}{3} = 2

a_T^2 + a_N^2 = 4 + 4 = 8

\|\mathbf{a}\|^2 = 0 + 4 + 4 = 8

Answer: a_T = 2 and a_N = 2. As a second check, the speed is t² + 2, whose derivative 2t equals 2 at t = 1, matching a_T.

Example 3: Constant speed on a helix

A particle moves along the helix r(t) = ⟨3 cos t, 3 sin t, 4t⟩. Find a_T and a_N.

\mathbf{v}(t) = \langle -3\sin t, \; 3\cos t, \; 4 \rangle, \quad \|\mathbf{v}(t)\| = 5

\mathbf{a}(t) = \langle -3\cos t, \; -3\sin t, \; 0 \rangle

\mathbf{v} \cdot \mathbf{a} = 9\sin t\cos t - 9\sin t\cos t + 0 = 0

a_T = \dfrac{0}{5} = 0

\|\mathbf{a}\| = \sqrt{9\cos^2 t + 9\sin^2 t} = 3

a_N = \sqrt{\|\mathbf{a}\|^2 - a_T^2} = \sqrt{9 - 0} = 3

\kappa\|\mathbf{v}\|^2 = \dfrac{3}{25}(25) = 3

Answer: a_T = 0, because the speed never changes, and a_N = 3. All of the acceleration goes into turning.

Example 4: From acceleration to position

An object has acceleration a(t) = ⟨6t, 2, 0⟩, initial velocity v(0) = ⟨1, 0, 3⟩ and initial position r(0) = ⟨0, 1, 0⟩. Find v(t), r(t) and r(2).

\mathbf{v}(t) = \langle 3t^2 + C_1, \; 2t + C_2, \; C_3 \rangle

\langle C_1, C_2, C_3 \rangle = \langle 1, 0, 3 \rangle

\mathbf{v}(t) = \langle 3t^2 + 1, \; 2t, \; 3 \rangle

\mathbf{r}(t) = \langle t^3 + t + D_1, \; t^2 + D_2, \; 3t + D_3 \rangle

\langle D_1, D_2, D_3 \rangle = \langle 0, 1, 0 \rangle

\mathbf{r}(t) = \langle t^3 + t, \; t^2 + 1, \; 3t \rangle

\mathbf{r}(2) = \langle 8 + 2, \; 4 + 1, \; 6 \rangle = \langle 10, 5, 6 \rangle

Answer: v(t) = ⟨3t² + 1, 2t, 3⟩, r(t) = ⟨t³ + t, t² + 1, 3t⟩, and at t = 2 the object is at (10, 5, 6).

Example 5: A kicked ball (word problem)

A soccer ball is kicked from the origin on level ground with initial velocity ⟨8, 6, 14.7⟩ m/s, where z is height. Gravity gives a = ⟨0, 0, −9.8⟩ m/s². Find where it lands, how far it travels along the ground, its maximum height, and its speed when it lands.

\mathbf{v}(t) = \langle 8, \; 6, \; 14.7 - 9.8t \rangle

\mathbf{r}(t) = \langle 8t, \; 6t, \; 14.7t - 4.9t^2 \rangle

14.7t - 4.9t^2 = 0

4.9t(3 - t) = 0

t = 3

\mathbf{r}(3) = \langle 24, \; 18, \; 0 \rangle

\sqrt{24^2 + 18^2} = \sqrt{576 + 324} = \sqrt{900} = 30

14.7 - 9.8t = 0 \Rightarrow t = 1.5

z(1.5) = 14.7(1.5) - 4.9(1.5)^2 = 22.05 - 11.025 = 11.025

\mathbf{v}(3) = \langle 8, \; 6, \; -14.7 \rangle

\|\mathbf{v}(3)\| = \sqrt{64 + 36 + 216.09} = \sqrt{316.09} \approx 17.78

Answer: the ball lands at (24, 18, 0) after 3 s, 30 m from the kick. It reaches about 11.0 m high at t = 1.5 s and lands at about 17.78 m/s, the same speed it was kicked with.

Common mistakes

Related methods

Projectile motion in the plane

When a projectile moves in one vertical plane, you can use two components, as in projectile motion with parametric equations. The 3D version in Example 5 is the same, with one more horizontal direction.

Normal acceleration and curvature

For circular motion of radius R at speed v, a_N = κv² = v²/R. This is the centripetal acceleration from physics. On a curve with curvature κ, the same formula a_N = κ‖v‖² holds at each point.

Motion in the plane

For motion in the xy-plane, write the vectors with a third component of 0 so the cross product formula for a_N still works, or use the two-component tools in vector-valued functions.

Practice problems

1. Find the velocity, speed and acceleration of r(t) = ⟨t, t², t³⟩ at t = 1.
2. Find the speed and the acceleration of r(t) = ⟨cos t, sin t, t⟩.
3. Find a_T and a_N for the helix in Problem 2.
4. Find a_T and a_N for r(t) = ⟨t, t², t³⟩ at t = 1.
5. An object has a(t) = ⟨0, 0, −10⟩, v(0) = ⟨3, 4, 20⟩ and r(0) = ⟨0, 0, 0⟩. Find v(t) and r(t).
6. When and where does the object in Problem 5 return to the ground (z = 0)?
7. Find the maximum height of the object in Problem 5.
8. Find r(t) if a(t) = ⟨2, 6t, 0⟩, v(0) = ⟨0, 0, 1⟩ and r(0) = ⟨1, 0, 0⟩.
9. A car drives around a circular track of radius 50 m at a constant 20 m/s. Find a_T and a_N.
10. At the moment the car in Problem 9 is going 20 m/s, it starts speeding up at 2 m/s². Find a_T, a_N and the size of the total acceleration.

\mathbf{v}(t) = \langle 1, \; 2t, \; 3t^2 \rangle, \quad \mathbf{v}(1) = \langle 1, 2, 3 \rangle

\|\mathbf{v}(1)\| = \sqrt{1 + 4 + 9} = \sqrt{14} \approx 3.74

\mathbf{a}(t) = \langle 0, \; 2, \; 6t \rangle, \quad \mathbf{a}(1) = \langle 0, 2, 6 \rangle

\mathbf{v}(t) = \langle -\sin t, \; \cos t, \; 1 \rangle

\|\mathbf{v}(t)\| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}

\mathbf{a}(t) = \langle -\cos t, \; -\sin t, \; 0 \rangle

\mathbf{v} \cdot \mathbf{a} = \sin t\cos t - \sin t\cos t + 0 = 0

a_T = 0

\|\mathbf{a}\| = 1, \quad a_N = \sqrt{1 - 0} = 1

\mathbf{v} \cdot \mathbf{a} = (1)(0) + (2)(2) + (3)(6) = 22

a_T = \dfrac{22}{\sqrt{14}} \approx 5.880

\mathbf{v} \times \mathbf{a} = \langle 2(6) - 3(2), \; 3(0) - 1(6), \; 1(2) - 2(0) \rangle = \langle 6, -6, 2 \rangle

\|\mathbf{v} \times \mathbf{a}\| = \sqrt{36 + 36 + 4} = \sqrt{76}

a_N = \dfrac{\sqrt{76}}{\sqrt{14}} \approx 2.330

a_T^2 + a_N^2 = \dfrac{484}{14} + \dfrac{76}{14} = 40 = \|\mathbf{a}\|^2

\mathbf{v}(t) = \langle 3, \; 4, \; 20 - 10t \rangle

\mathbf{r}(t) = \langle 3t, \; 4t, \; 20t - 5t^2 \rangle

20t - 5t^2 = 0

5t(4 - t) = 0

t = 4

\mathbf{r}(4) = \langle 12, \; 16, \; 0 \rangle

20 - 10t = 0

t = 2

z(2) = 20(2) - 5(2)^2 = 40 - 20 = 20

\mathbf{v}(t) = \langle 2t, \; 3t^2, \; 1 \rangle

\mathbf{r}(t) = \langle t^2 + 1, \; t^3, \; t \rangle

a_T = 0

a_N = \kappa v^2 = \dfrac{1}{50}(20)^2

= \dfrac{400}{50} = 8

a_T = 2, \quad a_N = 8

\|\mathbf{a}\| = \sqrt{2^2 + 8^2} = \sqrt{68} \approx 8.25

Motion in space FAQ

How do you find velocity and acceleration from a position vector?

Differentiate once for velocity, v(t) = r′(t), and again for acceleration, a(t) = r″(t). The speed is the length of the velocity, ‖v(t)‖.

What are the tangential and normal components of acceleration?

They are the parts of a along T and N: a_T = (v · a)/‖v‖ changes the speed, and a_N = ‖v × a‖/‖v‖ changes the direction. Together, a = a_T T + a_N N.

Why is the tangential component 0 at constant speed?

a_T is the derivative of the speed, d‖v‖/dt. If the speed does not change, that derivative is 0, and all of the acceleration is normal.

How do you find position from acceleration?

Integrate a(t) to get v(t) plus a constant vector, use v(0) to find it, then integrate v(t) and use r(0). Each integration needs its own initial condition.

Summary

• Velocity v = r′, speed ‖v‖, acceleration a = v′ = r″.
• a = a_T T + a_N N, with a_T = (v · a)/‖v‖ and a_N = ‖v × a‖/‖v‖ = κ‖v‖².
• a_T changes the speed (it can be negative); a_N changes the direction and is never negative.
• Check with ‖a‖² = a_T^2 + a_N^2.
• Integrate a, then v, with initial conditions, to recover the path; for gravity, a = ⟨0, 0, −9.8⟩ m/s².

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