Rectilinear Motion

If the position of a particle at time $t$ is given by $s (t)$ , then its velocity is v(t) = s'(t) and its acceleration is a(t) = v'(t) = s''(t). Integration allows us to reverse this process: from acceleration, we recover velocity, and from velocity, we recover position.

Quantity	Relation	Direction
Position $s (t)$	Differentiate once to get $v (t)$	$s \to v$
Velocity v(t) = s'(t)	Differentiate once to get $a (t)$	$v \to a$
Acceleration a(t) = v'(t)	Integrate once to get $v (t)$	$a \to v$
Velocity $v (t)$	Integrate once to get $s (t)$	$v \to s$

If the position of a particle at time $t$ is given by $s (t)$ , then its velocity $v (t)$ is

v(t) = s'(t)

and its acceleration $a (t)$ is

a(t) = v'(t) = s''(t).

Conversely, the reverse process is fundamental in physics. According to Newton's second law of motion ( $F = m a$ ), if the net force $F$ acting on a particle of mass $m$ is known, its acceleration can be determined by $a = F / m$ . Because forces are readily modeled in many physical systems, we can use this relationship to find the velocity and position of a particle at any time, provided its initial position and initial velocity are known. This technique is also used in the navigation systems of submarines and spacecraft.

Find the equation describing the position of a particle if it moves in a straight line with constant acceleration, knowing the velocity and position of the particle at $t = 0$ .

Solution

Because acceleration is a constant, say $a_{0}$ , we have v'(t) = a_0. Integrating,

v (t) = \int a_{0} d t = a_{0} t + C .

To determine $C$ , suppose the velocity is $v_{0}$ at $t = 0$ ; that is, let $v (0) = v_{0}$ . Therefore

v_{0} = a_{0} (0) + C \Rightarrow C = v_{0},

and

\begin{aligned} v(t) &= a_0 t + C \\ &= a_0 t + v_0. \end{aligned}

Since s'(t) = v(t), we have s'(t) = a_0 t + v_0. Integrating gives us

\begin{aligned} s(t) &= \int(a_0 t + v_0)\,dt \\ &= \frac{1}{2}a_0 t^2 + v_0 t + C_1. \end{aligned}

To determine $C_{1}$ , suppose the position is $s_{0}$ at $t = 0$ ; that is, let $s (0) = s_{0}$ . Therefore,

s_{0} = 0 + 0 + C_{1} \Rightarrow C_{1} = s_{0},

and

s (t) = \frac{1}{2} a_{0} t^{2} + v_{0} t + s_{0} .

The force of gravitation exerted on an object is the weight of the object, $m g$ , where $m$ is the mass of the object and $g$ is the acceleration due to gravity. For objects near the surface of the earth, we can assume that $g$ is a constant ( $g = 9.8$ m/s² or $g = 32$ ft/s², approximately).

If the positive direction is taken to be downward, then the gravitational acceleration is a positive constant; that is, $a (t) = a_{0} = g$ . In this case, the equations describing the motion of a stone in vacuum falling from rest ( $v_{0} = 0$ ) are $v (t) = g t, s (t) = \frac{1}{2} g t^{2},$
assuming $s = 0$ at $t = 0$ (i.e., $s_{0} = 0$ ). We may eliminate $t$ between the above equations and find a relation between the position of the stone and its velocity at each moment:
$v (t) = \sqrt{2 g s (t)} .$
If the object is initially thrown upward, we have $v_{0} < 0$ .
If the positive direction is taken to be upward, then $a (t) = a_{0} = - g$ and if the object is initially thrown upward we have $v_{0} > 0$ .

A ball is thrown upward from the edge of a building 60 ft above the ground with an initial velocity of 20 ft/s. Find the position and velocity of the ball at time $t$ , and determine when it hits the ground.

Solution

Taking the positive direction to be upward, the acceleration due to gravity is

a (t) = - g = - 32 {ft/s}^{2} .

The initial conditions are $v_{0} = 20$ ft/s and $s_{0} = 60$ ft. From the result of the previous example,

v (t) = - 32 t + 20,

s (t) = - 16 t^{2} + 20 t + 60.

The ball hits the ground when $s (t) = 0$ :

- 16 t^{2} + 20 t + 60 = 0 ⟹ 4 t^{2} - 5 t - 15 = 0.

Applying the quadratic formula,

t = \frac{5 \pm \sqrt{25 + 240}}{8} = \frac{5 \pm \sqrt{265}}{8} .

Since $t \geq 0$ , we take the positive root:

t = \frac{5 + \sqrt{265}}{8} \approx 2.66 s .

A particle moves along a straight line with acceleration $a (t) = 6 t - 4$ . At time $t = 0$ , its velocity is $v_{0} = - 3$ m/s and its position is $s_{0} = 2$ m.

(a) Find the position of the particle at time $t$ .

(b) Determine all times at which it is momentarily at rest.

Solution

(a) Since v'(t) = a(t) = 6t - 4, we integrate to find the velocity:

v (t) = \int (6 t - 4) d t = 3 t^{2} - 4 t + C .

Applying the initial condition $v (0) = - 3$ :

- 3 = 0 - 0 + C ⟹ C = - 3,

v (t) = 3 t^{2} - 4 t - 3.

Since s'(t) = v(t), we integrate again:

s (t) = \int (3 t^{2} - 4 t - 3) d t = t^{3} - 2 t^{2} - 3 t + C_{1} .

Applying $s (0) = 2$ :

2 = 0 - 0 - 0 + C_{1} ⟹ C_{1} = 2,

and therefore

s (t) = t^{3} - 2 t^{2} - 3 t + 2.

(b) The particle is momentarily at rest when $v (t) = 0$ ; that is, when

3 t^{2} - 4 t - 3 = 0.

By the quadratic formula,

t = \frac{4 \pm \sqrt{16 + 36}}{6} = \frac{4 \pm \sqrt{52}}{6} = \frac{2 \pm \sqrt{13}}{3} .

Since $\sqrt{13} \approx 3.61$ , the two values are

t \approx \frac{2 + 3.61}{3} \approx 1.87 s and t \approx \frac{2 - 3.61}{3} \approx - 0.54 s .

Only the positive value is physically meaningful (assuming motion begins at $t = 0$ ), so the particle is at rest at $t \approx 1.87$ s.

Displacement and Total Distance Traveled

Two different questions can be asked about a moving particle: where did it end up, and how far did it actually travel? These give different answers whenever the particle changes direction.

The displacement of a particle over the time interval $[t_{1}, t_{2}]$ is the net change in its position,

Δ s = s (t_{2}) - s (t_{1}) .

It is a signed quantity: positive if the particle finishes to the positive side of where it started, negative if it finishes to the negative side, and zero if it returns to its starting point.

The total distance traveled is the length of the whole path covered, counting every leg of the trip as a positive amount, regardless of direction.

Displacement only compares the two endpoints; it has no memory of what happened in between. Total distance keeps a running tally of the path. If the particle never reverses direction, the two agree in size, and the total distance is simply $| Δ s |$ . As soon as the particle turns around, the trip out and the trip back partly cancel in the displacement but add in the distance, so

total distance ≥ |displacement|.

How to compute total distance.

Find the position function $s (t)$ by integrating the velocity.
Find the turning points: the times in $[t_{1}, t_{2}]$ where $v (t) = 0$ and the velocity changes sign.
Split $[t_{1}, t_{2}]$ at those times, and add the absolute change in position over each piece: $distance = | s (c_{1}) - s (t_{1}) | + | s (c_{2}) - s (c_{1}) | + \dots + | s (t_{2}) - s (c_{k}) | .$

The velocity vanishing at a time is not by itself enough; the velocity must actually change sign there for the particle to reverse direction.

A useful sanity check: the average velocity over $[t_{1}, t_{2}]$ uses the displacement, while the average speed uses the total distance. They are generally not the same number.

A ball is thrown straight up from ground level with an initial speed of 48 ft/s, so its velocity is $v (t) = - 32 t + 48$ ft/s. Find the displacement and the total distance traveled during the first 2 seconds.

Solution

Integrating the velocity with $s (0) = 0$ ,

s (t) = \int (- 32 t + 48) d t = - 16 t^{2} + 48 t .

The displacement over $[0, 2]$ is

s (2) - s (0) = (- 64 + 96) - 0 = 32 ft .

For the total distance, first locate the turning point. The velocity is zero when

- 32 t + 48 = 0 ⟹ t = 1.5 s,

and it changes from positive to negative there, so the ball rises until $t = 1.5$ s and falls afterward. Its height at the turning point is

s (1.5) = - 16 (2.25) + 48 (1.5) = 36 ft .

Splitting the interval at $t = 1.5$ ,

\begin{aligned} \text{distance} &= |s(1.5) - s(0)| + |s(2) - s(1.5)| \\ &= |36 - 0| + |32 - 36| \\ &= 36 + 4 = 40 \text{ ft}. \end{aligned}

The ball climbed 36 ft and then dropped back 4 ft, covering 40 ft of path while ending only 32 ft above its starting point. Correspondingly, its average velocity over the 2 seconds is $32 / 2 = 16$ ft/s, while its average speed is $40 / 2 = 20$ ft/s.

Exercises

A car traveling at 60 ft/s brakes with constant deceleration and comes to a stop after covering 120 ft.

(a) Find the (constant) acceleration.

(b) How long does the car take to stop?

Solution

(a) Take the direction of motion as positive, with $s_{0} = 0$ and $v_{0} = 60$ ft/s. With constant acceleration $a_{0}$ ,

v (t) = a_{0} t + 60, s (t) = \frac{1}{2} a_{0} t^{2} + 60 t .

The car stops at the time $t_{1}$ where $v (t_{1}) = 0$ , so $t_{1} = - 60 / a_{0}$ . Substituting this into the position equation and setting $s (t_{1}) = 120$ :

\begin{aligned} 120 &= \frac{1}{2}a_0\left(\frac{-60}{a_0}\right)^2 + 60\left(\frac{-60}{a_0}\right) \\ &= \frac{1800}{a_0} - \frac{3600}{a_0} \\ &= \frac{-1800}{a_0}. \end{aligned}

Hence $a_{0} = - 15$ ft/s²; the car decelerates at 15 ft/s².

(b) $t_{1} = - 60 / a_{0} = 60 / 15 = 4$ s.

A stone is dropped from rest at the top of a cliff and reaches the ground 3 s later. Taking $g = 9.8$ m/s² and ignoring air resistance, find the height of the cliff and the speed of the stone on impact.

Solution

Take downward as the positive direction, so $a (t) = g = 9.8$ , $v_{0} = 0$ , and $s_{0} = 0$ . Then

v (t) = 9.8 t, s (t) = 4.9 t^{2} .

At $t = 3$ s,

s (3) = 4.9 (9) = 44.1 m, v (3) = 9.8 (3) = 29.4 m/s .

The cliff is 44.1 m high and the stone strikes the ground at 29.4 m/s. As a check, the relation $v = \sqrt{2 g s}$ gives $\sqrt{2 (9.8) (44.1)} = \sqrt{864.36} = 29.4$ m/s.

A particle moves along a line with acceleration $a (t) = 12 t^{2} - 6$ . Its initial velocity is $v (0) = 2$ and its initial position is $s (0) = - 1$ . Find $v (t)$ and $s (t)$ .

Solution

Integrating the acceleration,

v (t) = \int (12 t^{2} - 6) d t = 4 t^{3} - 6 t + C .

Since $v (0) = 2$ , we get $C = 2$ , so

v (t) = 4 t^{3} - 6 t + 2.

Integrating again,

s (t) = \int (4 t^{3} - 6 t + 2) d t = t^{4} - 3 t^{2} + 2 t + C_{1} .

Since $s (0) = - 1$ , we get $C_{1} = - 1$ , and therefore

s (t) = t^{4} - 3 t^{2} + 2 t - 1.

A particle moves with acceleration $a (t) = \cos t$ , with $v (0) = 0$ and $s (0) = 1$ . Find its position at time $t$ , and find the largest and smallest values the position ever takes.

Solution

Integrating,

v (t) = \int \cos t d t = \sin t + C .

From $v (0) = 0$ we get $C = 0$ , so $v (t) = \sin t$ . Integrating once more,

s (t) = \int \sin t d t = - \cos t + C_{1} .

From $s (0) = 1$ we have $- 1 + C_{1} = 1$ , so $C_{1} = 2$ and

s (t) = 2 - \cos t .

Because $- 1 \leq \cos t \leq 1$ , the position satisfies $1 \leq s (t) \leq 3$ . The particle oscillates between $s = 1$ and $s = 3$ , never leaving that interval.

A ball is thrown straight up from ground level with an initial speed of 48 ft/s. Using $g = 32$ ft/s², find the maximum height it reaches and the total time it stays in the air.

Solution

Take upward as positive, so $a_{0} = - 32$ , $v_{0} = 48$ , and $s_{0} = 0$ . Then

v (t) = - 32 t + 48, s (t) = - 16 t^{2} + 48 t .

The ball reaches its highest point when the velocity is zero:

- 32 t + 48 = 0 ⟹ t = 1.5 s .

The maximum height is

s (1.5) = - 16 (2.25) + 48 (1.5) = - 36 + 72 = 36 ft .

The ball returns to the ground when $s (t) = 0$ :

- 16 t^{2} + 48 t = - 16 t (t - 3) = 0 ⟹ t = 0 or t = 3.

So the ball is in the air for 3 s, exactly twice the time it took to reach the top.

A particle starts at the origin and moves with velocity $v (t) = t^{2} - 4 t + 3$ m/s. Find its displacement and the total distance traveled during the first 3 seconds, and explain why the two answers differ.

Solution

Integrating the velocity with $s (0) = 0$ ,

s (t) = \int (t^{2} - 4 t + 3) d t = \frac{t^{3}}{3} - 2 t^{2} + 3 t + C, C = 0.

The displacement over $[0, 3]$ is

s (3) - s (0) = (9 - 18 + 9) - 0 = 0 m .

For the total distance we must track the direction of motion. Factoring,

v (t) = t^{2} - 4 t + 3 = (t - 1) (t - 3),

so the velocity is positive on $(0, 1)$ and negative on $(1, 3)$ ; the particle reverses direction at $t = 1$ . Its position there is

s (1) = \frac{1}{3} - 2 + 3 = \frac{4}{3} m .

Adding the two legs of the trip,

| s (1) - s (0) | + | s (3) - s (1) | = \frac{4}{3} + \frac{4}{3} = \frac{8}{3} \approx 2.67 m .

The displacement is zero because the particle ends where it started, while the total distance measures the whole path traveled, out and back.

Rectilinear Motion