Motion in a Straight Line (College Board AP® Calculus AB)

Revision Note

Author

Jamie Wood

Expertise

Maths

Introduction to straight line motion

What is straight line motion?

Straight line motion models how objects move in a straight line, with respect to time
- This may be described as motion 'along the $x$ -axis'
- The straight line will have a positive and a negative direction
  - On the $x$ -axis this will be the usual positive and negative directions
  - If the question doesn't specify, you can choose the positive and negative directions
  - (Just be consistent once you've made a choice!)

What terminology do I need to be aware of?

Particle
- A particle is the general term used for an object
- A particle is assumed to be the 'size' of a single point
  - So you don't need to worry about its 3D dimensions
Time $t$
- Time is usually measured in seconds ( $s$ )
- Displacement, velocity and acceleration are all functions of time $t$
- 'Initial' or 'Initially' means 'when $t = 0$ '
Displacement $s$
- $s$ is the usual notation for displacement
  - For motion along the $x$ -axis, $x$ may be used instead of $s$
- Displacement is usually measured in feet ( $ft$ ) or meters ( $m$ )
  - Sometimes questions may not state a particular unit, and will simply use "units"
- The displacement of a particle is its distance relative to a fixed point
  - The fixed point may be (but is not always) the particle’s initial position
- Displacement will be zero, $s = 0$ , when the object is at the fixed point
- Otherwise the displacement will be
  - positive if the particle is in the positive direction from the fixed point
  - or negative if it is in the negative direction from the fixed point

Distance $d$
- Distance is the magnitude of displacement
- Use of the word distance could refer to
  - the distance traveled by a particle
  - the (straight line) distance the particle is from a particular point
- Be careful not to confuse displacement with distance
  - Consider a bus starting and ending its journey at a bus depot,
  - its displacement will be zero when it returns to the depot
  - but the distance the bus has travelled will be the length of the route
- Distance is always positive (or zero)
Velocity $v$
- Velocity is usually measured in feet or meters per second
- The velocity of a particle is the rate of change of its displacement at time $t$
  - Velocity will be positive if the particle is moving in the positive direction
  - Or negative if it is moving in the negative direction
- If the particle is stationary, that means the velocity is zero, $v = 0$
  - '(Instantaneously) at rest' also means that $v = 0$
Speed
- Speed is the magnitude (i.e. absolute value or modulus) of the velocity
- For a particle moving in a straight line
  - speed is the 'velocity ignoring the direction'
  - if $v = 4$ , speed $= |4| = 4$
  - if $v = - 6$ , speed = $|- 6| = 6$
Acceleration $a$
- Acceleration is usually measured in feet or meters per second squared
  - That is the same as feet or meters per second per second
- The acceleration of a particle is the rate of change of its velocity at time $t$
- Acceleration can be positive or negative
  - but the sign alone cannot fully describe the particle’s motion
- If velocity and acceleration have the same sign
  - then the particle is accelerating (speeding up)
- if velocity and acceleration have different signs
  - then the particle is decelerating (slowing down)
- At times when the acceleration is zero, $a = 0$ ,
  - the particle is moving with constant velocity
- In all cases the direction of motion is determined by the sign (+ or -) of the velocity
- There is no special term for the magnitude of acceleration
  - "The magnitude of the acceleration" is simply used instead

Velocity & acceleration as derivatives

What is velocity as a derivative?

Velocity is the rate of change of displacement
- $v = \frac{d s}{d t}$
- (differentiate displacement to get velocity)
Velocity is the slope of a displacement-time graph

What is acceleration as a derivative?

Acceleration is the rate of change of velocity
- $a = \frac{d v}{d t}$
- (differentiate velocity to get acceleration)
Acceleration is the slope of a velocity-time graph
This means that acceleration is also the rate of change, of the rate of change of displacement
- $a = \frac{d^{2} s}{d t^{2}}$
- (differentiate displacement twice to get acceleration)

Worked Example

The displacement from the origin of a particle, $P$ , as it travels along the $x$ -axis is given by $s_{P} = 3 \sin (\frac{1}{2} t)$ .

The displacement from the origin of a second particle, $Q$ , as it travels along the $x$ -axis is given by $s_{Q} = \frac{1}{8} t^{3} - \frac{7}{4} t^{2} + 5 t$ .

$s_{P}$ and $s_{Q}$ are measured in meters and $t$ is measured in seconds for $0 \leq t \leq 10$ .

(a) Determine which particle is furthest from the origin at $t = 5$ .

Answer:

Find the displacement of each particle at $t = 5$

$s_{P} = 3 \sin (\frac{1}{2} \cdot 5) = 1.795416 . . .$ meters

$s_{Q} = \frac{1}{8} {(5)}^{3} - \frac{7}{4} {(5)}^{2} + 5 (5) = - 3.125$ meters

$P$ is 1.795... meters away from the origin in the positive direction, while $Q$ is 3.125 meters away from the origin in the negative direction

At $t = 5$ particle $Q$ is the furthest from the origin.

(b) At $t = 5$ , determine if the particles are moving closer together, or further apart. Explain your reasoning in your working.

Answer:

This question is about the direction of motion, so we need to find the velocity

Find expressions for the velocities by differentiating the expressions for the displacements

$v_{P} = \frac{d s_{P}}{d t} = \frac{3}{2} \cos (\frac{1}{2} t)$

$v_{Q} = \frac{d s_{Q}}{d t} = \frac{3}{8} t^{2} - \frac{7}{2} t + 5$

Find the velocities when $t = 5$

$v_{P} = \frac{3}{2} \cos (\frac{1}{2} \cdot 5) = - 1.201715 . . .$ meters per second

$v_{Q} = \frac{3}{8} {(5)}^{2} - \frac{7}{2} (5) + 5 = - 3.125$ meters per second

Consider the positions and velocities of the two particles

$P$ is on the positive side (right) of the $x$ axis (1.795... meters), moving with a negative velocity, so back towards the origin (moving to the left)

$Q$ is on the negative side (left) of the $x$ axis (-3.125... meters), moving with a negative velocity, so further away from the origin (moving to the left)

$\overset{3.125}{\leftarrow} \overset{1.2017}{\leftarrow} Q P$

So both particles are moving to the left (negative $x$ direction) but $Q$ is moving faster, and its position is further to the left, so $Q$ is "escaping" from $P$

Therefore at $t = 5$ , the particles are moving further apart

Answer:

Find expressions for the accelerations by differentiating the expressions for the velocities

$a_{P} = \frac{d v_{P}}{d t} = - \frac{3}{4} \sin (\frac{1}{2} t)$

$a_{Q} = \frac{d v_{Q}}{d t} = \frac{3}{4} t - \frac{7}{2}$

Find the acceleration of each when $t = 5$

$a_{P} = - \frac{3}{4} \sin (\frac{1}{2} \cdot 5) = - 0.448854 . . .$ meters per second squared

$a_{Q} = \frac{3}{4} (5) - \frac{7}{2} = 0.25$ meters per second squared

Whilst the acceleration of $Q$ is the largest (as it is positive whereas $P$ 's is negative), it is $P$ whose acceleration has the greatest absolute value (magnitude)

At $t = 5$ , particle $P$ has the greatest magnitude of acceleration.

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Motion in a Straight Line (College Board AP® Calculus AB)

Revision Note

Introduction to straight line motion

What is straight line motion?

What terminology do I need to be aware of?

Velocity & acceleration as derivatives

What is velocity as a derivative?

What is acceleration as a derivative?

Worked Example

You've read 0 of your 10 free revision notes

Unlock more, it's free!

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Author: Jamie Wood