Rectilinear Motion

When an object moves along a straight line, its position, velocity, and acceleration are all connected by differentiation. Velocity is the derivative of position with respect to time, and acceleration is the derivative of velocity.

Quantity	Definition	Formula
Average velocity	Change in position divided by time elapsed	$v_{avg} = \frac{Δ s}{Δ t}$
Instantaneous velocity	Derivative of position	$v (t) = \frac{d s}{d t}$
Average acceleration	Change in velocity divided by time elapsed	$a_{avg} = \frac{Δ v}{Δ t}$
Instantaneous acceleration	Derivative of velocity	$a (t) = \frac{d v}{d t} = \frac{d^{2} s}{d t^{2}}$
Speed	Absolute value of velocity	$speed = \| v (t) \|$

Setting Up the Problem

Consider an object that is moving along a straight line. Choose one direction as positive and the opposite as negative, and one point as the origin $O$ . Let $s$ be the object's position on this straight line, and let $s = f (t)$ denote the position of the object at time $t$ .

A number line with origin O, positive direction to the right, and a point marking the position s of the moving object

The object's average velocity during an interval of time $[t_{0}, t_{1}]$ is

v_{avg} = \frac{f (t_{1}) - f (t_{0})}{t_{1} - t_{0}} = \frac{Δ s}{Δ t} .

The instantaneous velocity at $t_{0}$ is obtained by letting the time interval shrink to zero:

v (t_{0}) = lim_{Δ t \to 0} \frac{Δ s}{Δ t}

and in general

v (t) = \frac{d s}{d t} and v (t_{0}) = {\frac{d s}{d t} |}_{t = t_{0}} .

Similarly, the average acceleration and instantaneous acceleration at time $t$ are

a_{avg} = \frac{Δ v}{Δ t}, and a (t) = \frac{d v}{d t} = \frac{d^{2} s}{d t^{2}} .

We often drop "instantaneous" and simply say velocity and acceleration instead of instantaneous velocity and instantaneous acceleration.
Speed is the absolute value of velocity. Therefore, speed is always nonnegative: $speed = | v (t) | = | \frac{d s}{d t} | .$
In general, average speed is not the absolute value of the average velocity: $average speed \neq | v_{avg} |$ . The average speed is the total distance traveled divided by the travel time. For example, if you travel to a city 150 km away and return in 4 hours, your average speed is $(150 + 150) / 4 = 75$ km/h, but your average velocity is zero.

Examples

Suppose that the position $s$ of a body at time $t$ is given by $s = 150 + 30 t - 16 t^{2}$ , where $s$ is measured in feet and $t$ in seconds. Determine:

the average velocity in the first three seconds $0 \leq t \leq 3$ s;
the velocity at $t = 3$ s;
the average acceleration in the first 3 seconds;
the acceleration at $t = 3$ s.

Solution

(a) To calculate the average velocity, first find

Δ s

: \begin{aligned} \Delta s &= s(3) - s(0) \\ &= \left(150 + 30(3) - 16(9)\right) - \left(150\right) \\ &= 90 - 144 = -54. \end{aligned} Therefore, the average velocity is

v_{avg} = \frac{Δ s}{Δ t} = - \frac{54}{3} = - 18 ft/s.

(b) The instantaneous velocity is

s = 150 + 30 t - 16 t^{2} ⟹ v (t) = \frac{d s}{d t} = 30 - 32 t .

The velocity at

t = 3

s is

v (3) = 30 - 32 (3) = - 66 ft/s.

(c) To determine the average acceleration, calculate

Δ v

: \begin{aligned} \Delta v &= v(3) - v(0) \\ &= (30 - 96) - 30 \\ &= -96 \text{ ft/s.} \end{aligned} The average acceleration is

a_{avg} = \frac{Δ v}{Δ t} = - \frac{96}{3} = - 32 {ft/s}^{2} .

(d) The acceleration is

a (t) = \frac{d v}{d t} = - 32 {ft/s}^{2} .

In this example the acceleration is constant, so the average acceleration equals the instantaneous acceleration at every time.

The position of a particle is given by $s = A \sin (ω t)$ , where $A$ and $ω$ are constants. Find the velocity and acceleration of the particle.

Solution

Using the Chain Rule:

s = A \sin ω t

\begin{aligned} v &= \frac{ds}{dt} \\ &= A \frac{d(\sin(\omega t))}{d(\omega t)} \cdot \frac{d(\omega t)}{dt} \\ &= A\cos(\omega t)\cdot\omega \\ &= A\omega\cos(\omega t). \end{aligned} And the acceleration: \begin{aligned} a &= \frac{dv}{dt} \\ &= A\omega \frac{d(\cos(\omega t))}{d(\omega t)} \cdot \frac{d(\omega t)}{dt} \\ &= A\omega(-\sin(\omega t))(\omega) \\ &= -A\omega^2\sin(\omega t). \end{aligned}

Frequently Asked Questions

What does negative velocity mean physically?

Negative velocity means the object is moving in the negative direction (opposite to the chosen positive direction). For example, if rightward is positive, negative velocity means the object is moving left.

What is the difference between speed and velocity?

Velocity is a signed quantity: it records both how fast the object moves and in which direction. Speed is the magnitude of velocity, so it is always nonnegative. An object can have constant speed while its velocity changes direction (as in circular motion).

When is an object momentarily at rest?

An object is momentarily at rest when

v (t) = 0

. This is also the condition for a potential turning point in rectilinear motion, where the object might reverse direction.

What does the sign of acceleration tell us?

Positive acceleration means velocity is increasing; negative acceleration means velocity is decreasing. Note that negative acceleration does not necessarily mean the object is slowing down: if the velocity is also negative, the object is actually speeding up (in the negative direction).

Setting Up the Problem

Examples

Frequently Asked Questions

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