Velocity and Acceleration

Section 3.1 Velocity and Acceleration

Subsection 3.1.1 Velocity and Acceleration

If you are moving along the

x

–axis and your position at time

t

x (t),

then your velocity at time

t

v (t) = x^{'} (t)

and your acceleration at time

t

a (t) = v^{'} (t) = x^{″} (t) .

Your speed, i.e. the distance that you are travelling per unit time, is

| v (t) | = | x^{'} (t) | .

Example 3.1.1. Velocity as derivative of position.

Suppose that you are moving along the

x

–axis and that at time

t

your position is given by

\begin{aligned} x (t) & = t^{3} - 3 t + 2. \end{aligned}

We’re going to try and get a good picture of what your motion is like. We can learn quite a bit just by looking at the sign of the velocity

v (t) = x^{'} (t)

at each time

t .

If $x^{'} (t) > 0,$ then at that instant $x$ is increasing, i.e. you are moving to the right.
If $x^{'} (t) = 0,$ then at that instant you are not moving at all.
If $x^{'} (t) < 0,$ then at that instant $x$ is decreasing, i.e. you are moving to the left.

From the given formula for

x (t)

it is straight forward to work out the velocity

\begin{aligned} v (t) = x^{'} (t) & = 3 t^{2} - 3 = 3 (t^{2} - 1) = 3 (t + 1) (t - 1) \end{aligned}

This is zero only when

t = - 1

and when

t = + 1;

at no other value

This is because the equation

a b = 0

is only satisfied for real numbers

a

and

b

when either

a = 0

b = 0

or both

a = b = 0 .

Hence if a polynomial is the product of two (or more) factors, then it is only zero when at least one of those factors is zero. There are more complicated mathematical environments in which you have what are called “zero divisors” but they are beyond the scope of this course.

t

can this polynomial be equal zero. Consequently in any time interval that does not include either

t = - 1

t = + 1,

v (t)

takes only a single sign

This is because if

v (t_{a}) < 0

and

v (t_{b}) > 0

then, by the intermediate value theorem, the continuous function

v (t) = x^{'} (t)

must take the value

0

for some

t

between

t_{a}

and

t_{b} .

. So

For all $t < - 1,$ both $(t + 1)$ and $(t - 1)$ are negative (sub in, for example, $t = - 10$ ) so the product $v (t) = x^{'} (t) = 3 (t + 1) (t - 1) > 0 .$
For all $- 1 < t < 1,$ the factor $(t + 1) > 0$ and the factor $(t - 1) < 0$ (sub in, for example, $t = 0$ ) so the product $v (t) = x^{'} (t) = 3 (t + 1) (t - 1) < 0 .$
For all $t > 1,$ both $(t + 1)$ and $(t - 1)$ are positive (sub in, for example, $t = + 10$ ) so the product $v (t) = x^{'} (t) = 3 (t + 1) (t - 1) > 0 .$

The figure below gives a summary of the sign information we have about

t - 1,

t + 1

and

x^{'} (t) .

It is now easy to put together a mental image of your trajectory.

For $t$ large and negative (i.e. far in the past), $x (t)$ is large and negative and $v (t)$ is large and positive. For example
³
Notice here we are using the fact that when $t$ is very large $t^{3}$ is much bigger than $t^{2}$ and $t^{1} .$ So we can approximate the value of the polynomial $x (t)$ by the largest term — in this case $t^{3} .$ We can do similarly with $v (t)$ — the largest term is $3 t^{2} .$
, when $t = - 10^{6},$ $x (t) \approx t^{3} = - 10^{18}$ and $v (t) \approx 3 t^{2} = 3 \cdot 10^{12} .$ So you are moving quickly to the right.
For $t < - 1,$ $v (t) = x^{'} (t) > 0$ so that $x (t)$ is increasing and you are moving to the right.
At $t = - 1,$ $v (- 1) = 0$ and you have come to a halt at position $x (- 1) = (- 1)^{3} - 3 (- 1) + 2 = 4 .$
For $- 1 < t < 1,$ $v (t) = x^{'} (t) < 0$ so that $x (t)$ is decreasing and you are moving to the left.
At $t = + 1,$ $v (1) = 0$ and you have again come to a halt, but now at position $x (1) = 1^{3} - 3 + 2 = 0 .$
For $t > 1,$ $v (t) = x^{'} (t) > 0$ so that $x (t)$ is increasing and you are again moving to the right.
For $t$ large and positive (i.e. in the far future), $x (t)$ is large and positive and $v (t)$ is large and positive. For example
⁴
We are making a similar rough approximation here.
, when $t = 10^{6},$ $x (t) \approx t^{3} = 10^{18}$ and $v (t) \approx 3 t^{2} = 3 \cdot 10^{12} .$ So you are moving quickly to the right.

Here is a sketch of the graphs of

x (t)

and

v (t) .

The heavy lines in the graphs indicate when you are moving to the right — that is where

v (t) = x^{'} (t)

is positive.

And here is a schematic picture of the whole trajectory.

Example 3.1.2. Position and velocity from acceleration.

In this example we are going to figure out how far a body falling from rest will fall in a given time period.

We should start by defining some variables and their units. Denote
- time in seconds by $t,$
- mass in kilograms by $m,$
- distance fallen (in metres) at time $t$ by $s (t),$ velocity (in m/sec) by $v (t) = s^{'} (t)$ and acceleration (in m/sec $^{2}$ ) by $a (t) = v^{'} (t) = s^{″} (t) .$
It makes sense to choose a coordinate system so that the body starts to fall at $t = 0 .$
We will use Newton’s second law of motion
$\begin{matrix} the force applied to the body at time t = m \cdot a (t) . \end{matrix}$
together with the assumption that the only force acting on the body is gravity (in particular, no air resistance). Note that near the surface of the Earth,
$\begin{aligned} the force due to gravity acting on a body of mass m & = m \cdot g . \end{aligned}$
The constant $g,$ called the acceleration of gravity
⁵
It is also called the standard acceleration due to gravity or standard gravity. For those of you who prefer imperial units (or US customary units), it is about $32$ ft/sec $^{2},$ $77165$ cubits/minute $^{2},$ or $631353$ furlongs/hour $^{2} .$
, is about 9.8m/sec $^{2} .$
Since the body is falling from rest, we know that its initial velocity is zero. That is
$\begin{aligned} v (0) & = 0. \end{aligned}$
Newton’s second law then implies that
$\begin{aligned} m \cdot a (t) & = force due to gravity \\ m \cdot v^{'} (t) & = m \cdot g & cancel the m \\ v^{'} (t) & = g \end{aligned}$
In order to find the velocity, we need to find a function of $t$ whose derivative is constant. We are simply going to guess such a function and then we will verify that our guess has all of the desired properties. It’s easy to guess a function whose derivative is the constant $g .$ Certainly $g t$ has the correct derivative. So does

$\begin{matrix} v (t) = g t + c \end{matrix}$

for any constant $c .$ One can then verify
⁶
While it is clear that this satisfies the equation we want, it is less clear that it is the only function that works. To see this, assume that there are two functions $f (t)$ and $h (t)$ which both satisfy $v^{'} (t) = g .$ Then $f^{'} (t) = h^{'} (t) = g$ and so $f^{'} (t) - h^{'} (t) = 0 .$ Equivalently $\frac{d}{d t} (f (t) - h (t)) = 0.$ The only function whose derivative is zero everywhere is the constant function (see Section 2.13 and Corollary 2.13.12). Thus $f (t) - h (t) = constant .$ So all the functions that satisfy $v^{'} (t) = g$ must be of the form $g t + constant .$
that $v^{'} (t) = g .$ Using the fact that $v (0) = 0$ we must then have $c = 0$ and so

$\begin{aligned} v (t) & = g t . \end{aligned}$
Since velocity is the derivative of position, we know that
$\begin{aligned} s^{'} (t) & = v (t) = g \cdot t . \end{aligned}$
To find $s (t)$ we are again going to guess and check. It’s not hard to see that we can use
$\begin{aligned} s (t) & = \frac{g}{2} t^{2} + c \end{aligned}$
where again $c$ is some constant. Again we can verify that this works simply by differentiating
⁷
To show that any solution of $s^{'} (t) = g v$ must be of this form we can use the same reasoning we used to get $v (t) = g t + constant .$
. Since we have defined $s (t)$ to be the distance fallen, it follows that $s (0) = 0$ which in turn tells us that $c = 0 .$ Hence
$\begin{aligned} s (t) & = \frac{g}{2} t^{2} & but g = 9.8, so \\ = 4.9 t^{2}, \end{aligned}$
which is exactly the $s (t)$ used way back in Section 1.2.

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Let’s now do a similar but more complicated example.

Example 3.1.3. Stopping distance of a braking car.

A car’s brakes can decelerate the car at 64000

{km/hr}^{2} .

How fast can the car be driven if it must be able to stop within a distance of 50m?

Solution Before getting started, notice that there is a small “trick” in this problem — several quantities are stated but their units are different. The acceleration is stated in kilometres per hour

^{2},

but the distance is stated in metres. Whenever we come across a “real world” problem

⁸

Well — “realer world” would perhaps be a betterer term.

we should be careful of the units used.

We should first define some variables and their units. Denote
- time (in hours) by $t,$
- the position of the car (in kilometres) at time $t$ by $x (t),$ and
- the velocity (in kilometres per hour) by is $v (t) .$
We can also choose a coordinate system such that $x (0) = 0$ and the car starts braking at time $t = 0 .$
Now let us rewrite the information in the problem in terms of these variables.
- We are told that, at maximum braking, the acceleration $v^{'} (t) = x^{″} (t)$ of the car is $- 64000 .$
- We need to determine the maximum initial velocity $v (0)$ so that the stopping distance is at most $50 m = 0.05 k m$ (being careful with our units). Let us call the stopping distance $x_{s t o p}$ which is really $x (t_{s t o p})$ where $t_{s t o p}$ is the stopping time.
In order to determine $x_{s t o p}$ we first need to determine $t_{s t o p},$ which we will do by assuming maximum braking from a, yet to be determined, initial velocity of $v (0) = q$ m/sec.
Assuming that the car undergoes a constant acceleration at this maximum braking power, we have

$\begin{aligned} v^{'} (t) & = - 64000 \end{aligned}$

This equation is very similar to the ones we had to solve in Example 3.1.2 just above.

As we did there
⁹
Now is a good time to go back and have a read of that example.
, we are going to just guess $v (t) .$ First, we just guess one function whose derivative is $- 64000,$ namely $- 64000 t .$ Next we observe that, since the derivative of a constant is zero, any function of the form

$\begin{matrix} v (t) = - 64000 t + c \end{matrix}$

with constant $c,$ has the correct derivative. Finally, the requirement that the initial velocity $v (0) = q$ forces $c = q,$ so

$\begin{matrix} v (t) = q - 64000 t \end{matrix}$
From this we can easily determine the stopping time $t_{s t o p},$ when the initial velocity is $q,$ since this is just when $v (t) = 0 :$
$\begin{aligned} 0 = v (t_{s t o p}) & = q - 64000 \cdot t_{s t o p} & and so \\ t_{s t o p} & = \frac{q}{64000} . \end{aligned}$
Armed with the stopping time, how do we get at the stopping distance? We need to find the formula satisfied by $x (t) .$ Again (as per Example 3.1.2) we make use of the fact that
$\begin{aligned} x^{'} (t) & = v (t) = q - 64000 t . \end{aligned}$
So we need to guess a function $x (t)$ so that $x^{'} (t) = q - 64000 t .$ It is not hard to see that
$\begin{aligned} x (t) & = q t - 32000 t^{2} + constant \end{aligned}$
works. Since we know that $x (0) = 0,$ this constant is just zero and
$\begin{aligned} x (t) & = q t - 32000 t^{2} . \end{aligned}$
We are now ready to compute the stopping distance (in terms of the, still yet to be determined, initial velocity $q$ ):
$\begin{aligned} x_{s t o p} & = x (t_{s t o p}) = q t_{s t o p} - 32000 t_{s t o p}^{2} \\ = \frac{q^{2}}{64000} - \frac{32000 q^{2}}{64000^{2}} \\ = \frac{q^{2}}{64000} (1 - \frac{1}{2}) \\ = \frac{q^{2}}{2 \times 64000} \end{aligned}$
Notice that the stopping distance is a quadratic function of the initial velocity — if you go twice as fast, you need four times the distance to stop.
But we are told that the stopping distance must be less than $50 m = 0.05 k m .$ This means that
$\begin{aligned} x_{s t o p} = \frac{q^{2}}{2 \times 64000} & \leq \frac{5}{100} \\ q^{2} & \leq \frac{2 \times 64000 \times 5}{100} = \frac{64000 \times 10}{100} = 6400 \end{aligned}$
Thus we must have $q \leq 80 .$ Hence the initial velocity can be no greater than $80 k m / h .$

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Suppose you throw a ball straight up in the air, and its height from

t = 0

t = 4

is given by

h (t) = - 4.9 t^{2} + 19.6 t .

True or false: at time

t = 2,

the acceleration of the ball is 0.

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2.

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Suppose an object is moving with a constant acceleration. It takes ten seconds to accelerate from

1 \frac{m}{s}

2 \frac{m}{s} .

How long does it take to accelerate from

2 \frac{m}{s}

3 \frac{m}{s} ?

How long does it take to accelerate from

3 \frac{m}{s}

13 \frac{m}{s} ?

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3.

🔗

Let

s (t)

be the position of a particle at time

t .

True or false: if

s^{″} (a) > 0

for some

a,

then the particle’s speed is increasing when

t = a .

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4.

🔗

Let

s (t)

be the position of a particle at time

t .

True or false: if

s^{'} (a) > 0

and

s^{″} (a) > 0

for some

a,

then the particle’s speed is increasing when

t = a .

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Exercises — Stage 2 .

🔗

For this section, we will ask you a number of questions that have to do with objects falling on Earth. Unless otherwise stated, you should assume that an object falling through the air has an acceleration due to gravity of 9.8 meters per second per second.

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5.

🔗

A flower pot rolls out of a window 10m above the ground. How fast is it falling just as it smacks into the ground?

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6.

🔗

You want to know how deep a well is, so you drop a stone down and count the seconds until you hear it hit bottom.

If the stone took $x$ seconds to hit bottom, how deep is the well?
Suppose you think you dropped the stone down the well, but actually you tossed it down, so instead of an initial velocity of 0 metres per second, you accidentally imparted an initial speed of $1$ metres per second. What is the actual depth of the well, if the stone fell for $x$ seconds?

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7.

🔗

You toss a key to your friend, standing two metres away. The keys initially move towards your friend at 2 metres per second, but slow at a rate of 0.25 metres per second per second. How much time does your friend have to react to catch the keys? That is--how long are the keys flying before they reach your friend?

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8.

🔗

A car is driving at 100 kph, and it brakes with a deceleration of

50000 \frac{km}{{hr}^{2}} .

How long does the car take to come to a complete stop?

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9.

🔗

You are driving at 120 kph, and need to stop in 100 metres. How much deceleration do your brakes need to provide? You may assume the brakes cause a constant deceleration.

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10.

🔗

You are driving at 100 kph, and apply the brakes steadily, so that your car decelerates at a constant rate and comes to a stop in exactly 7 seconds. What was your speed one second before you stopped?

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11.

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About 8.5 minutes after liftoff, the US space shuttle has reached orbital velocity, 17 500 miles per hour. Assuming its acceleration was constant, how far did it travel in those 8.5 minutes?

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Source: https://www.nasa.gov/mission_pages/shuttle/shuttlemissions/sts121/launch/qa-leinbach.html

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12.

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A pitching machine has a dial to adjust the speed of the pitch. You rotate it so that it pitches the ball straight up in the air. How fast should the ball exit the machine, in order to stay in the air exactly 10 seconds?

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You may assume that the ball exits from ground level, and is acted on only by gravity, which causes a constant deceleration of 9.8 metres per second.

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13.

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A peregrine falcon can dive at a speed of 325 kph. If you were to drop a stone, how high up would you have to be so that the stone reached the same speed in its fall?

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14.

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You shoot a cannon ball into the air with initial velocity

v_{0},

and then gravity brings it back down (neglecting all other forces). If the cannon ball made it to a height of 100m, what was

v_{0} ?

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15.

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Suppose you are driving at 120 kph, and you start to brake at a deceleration of

50000

kph per hour. For three seconds you steadily increase your deceleration to

60000

kph per hour. (That is, for three seconds, the rate of change of your deceleration is constant.) How fast are you driving at the end of those three seconds?

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