Motion in a Plane · Lesson 5 of 13

Position, Velocity and Acceleration in a Plane

“Describe a curved trajectory using vector derivatives and explain the directions of velocity and acceleration.”

Learning Objectives

• Compute displacement and average velocity from position vectors. • Obtain instantaneous velocity and acceleration by differentiating components. • Distinguish a tangent velocity from the direction of acceleration. • Interpret speed and direction changes separately.

Track two coordinates with one clock

To locate an object moving on a tabletop, record x and y at each time. Combining them gives the position vector r = x i-hat + y j-hat. This vector points from the origin to the object. It is usually different from the object’s direction of motion. A particle can be northeast of the origin while travelling south. Position, velocity and acceleration answer different questions.

Displacement and average velocityLaTeX

Δ r = Δ x \hat{i} + Δ y \hat{j}, v_{av} = \frac{Δ r}{Δ t}

Average velocity points along the chord from the initial to final point on the path, when Δt > 0.

As the interval is shortened, the chord approaches the tangent to a smooth path. The direction of instantaneous velocity is along that tangent in the direction of motion, provided the speed is nonzero. At an instant of zero speed the zero velocity vector has no direction, though the path can still have a limiting tangent. The speed is the magnitude of the instantaneous velocity, not the magnitude of position divided by time.

Velocity componentsLaTeX

v = \frac{d r}{d t} = \frac{d x}{d t} \hat{i} + \frac{d y}{d t} \hat{j} = v_{x} \hat{i} + v_{y} \hat{j}, v = v_{x}^{2} + v_{y}^{2}

For fixed Cartesian axes, unit vectors are constant and differentiation acts on the components. Add dz/dt k-hat when needed in space.

Acceleration is the change of a vector

Average and instantaneous accelerationLaTeX

a_{av} = \frac{v _{f} - v _{i}}{Δ t}, a = \frac{d v}{d t} = a_{x} \hat{i} + a_{y} \hat{j}

The order is final velocity minus initial velocity. Acceleration points along the limiting change in velocity, not necessarily along the path.

Coordinate derivativesLaTeX

a_{x} = \frac{d v _{x}}{d t} = \frac{d ^{2} x}{d t ^{2}}, a_{y} = \frac{d v _{y}}{d t} = \frac{d ^{2} y}{d t ^{2}}

Speed and acceleration have magnitudes √(v_x² + v_y²) and √(a_x² + a_y²). Their directions require the component signs.

In one dimension, velocity and acceleration lie on one line. In a plane their angle can be anywhere from 0° to 180°. An acceleration component along velocity changes speed; a component perpendicular to velocity changes direction. An object following a curve can therefore accelerate while its speed stays constant. Conversely, knowing the acceleration alone does not determine the trajectory: the initial position and velocity also matter.

Turning through ninety degrees at the same speed

Problem
Velocity changes from 4 i-hat m/s to 4 j-hat m/s in 2 s. Find average acceleration.

1.The speed is 4 m/s at both endpoints, but the velocity vectors differ.
2.Change in velocity = −4 i-hat + 4 j-hat m/s.
3.Average acceleration = −2 i-hat + 2 j-hat m/s². Its magnitude is 2√2 = 2.83 m/s², directed 135° from +x.
4.Zero change in speed did not imply zero change in velocity. Endpoint data determine average acceleration but not its detailed variation during the interval.

A change in velocity when speed stays the same— The difference vector runs from the head of the initial velocity to the head of the final velocity. Equal arrow lengths mean equal speeds, not equal velocities.

Differentiate each coordinate independently

A position function with a constant height coordinate

Problem
Let r = 3t i-hat + 2t² j-hat + 5 k-hat metres. Find velocity and acceleration, and velocity at t = 1 s.

1.With t in seconds and coefficients carrying the required units, differentiation gives v = 3 i-hat + 4t j-hat m/s.
2.Differentiate again: a = 4 j-hat m/s². The constant z-coordinate has zero derivative, so there is no z-velocity or z-acceleration.
3.At 1 s, v = 3 i-hat + 4 j-hat m/s. Speed is 5 m/s and direction is arctan(4/3) = 53.13° above +x.
4.The particle remains in the plane z = 5 m. A nonzero k-component of position does not automatically mean three-dimensional motion.

A downward-bending trajectory

Problem
Let r = 3t i-hat − 2t² j-hat + 4 k-hat metres. Find velocity and acceleration at 2 s.

1.v = 3 i-hat − 4t j-hat m/s, so at 2 s it is 3 i-hat − 8 j-hat m/s.
2.a = −4 j-hat m/s². Speed at 2 s is √73 = 8.54 m/s.
3.Velocity points arctan(8/3) = 69.44° below positive x. The acceleration points vertically downward, at a different angle.
4.The constant z = 4 m again sets the plane of motion and contributes no velocity.

Distance and averages along a curved route

Average speed remains total path length divided by elapsed time. It can exceed the magnitude of average velocity even without a reversal on one axis: a curved path already adds more length than the straight chord. Instantaneous speed equals |v|. For a differentiable path, accumulated distance is the time integral of √(v_x² + v_y²), with a third squared component if the motion is in space. Integrating components of velocity instead yields vector displacement.

Quick check

A particle moves along a smooth curved path at constant nonzero speed. Which statement must be true?

Key Takeaways

• Position points from the origin; velocity points along the local direction of motion. • Differentiate each fixed-axis component to obtain velocity and acceleration. • Acceleration can change speed, direction or both. • Average acceleration uses final velocity minus initial velocity. • A constant speed can coexist with a changing velocity.

Previous · Lesson 4

Adding Vectors with Components

Next · Lesson 6

Motion in a Plane with Constant Acceleration

Track two coordinates with one clock

Acceleration is the change of a vector

Differentiate each coordinate independently

Distance and averages along a curved route

Quiz

Key Takeaways