15.2 Acceleration

15.16   Definition (Acceleration.) If a particle $\mbox{{\bf p}}$ moves in a straight line so that its position at time $t$ is $h(t)$, we have defined its velocity at time $t$ to be $h'(t)$. We now define its acceleration at time $t$ to be $h''(t)$, so that acceleration is the derivative of velocity. Thus if a particle moves with a constant acceleration of $1 \frac{\mbox{ft./sec.}}{\mbox{sec.}}$, then every second its velocity increases by one ft./sec.

15.17   Example. A mass on the end of a spring moves so that its height at time $t$ is $-A\cos(\omega t)$, where $A$ and $\omega$ are positive numbers. If we denote its velocity at time $t$ by $v(t)$, and its acceleration at time $t$ by $a(t)$ then

$$h(t) = -A\cos(\omega t) \mbox{{}}$$

$$v(t) = h'(t) = A\omega \sin(\omega t) \mbox{{}}$$

$$a(t) = v'(t) = A\omega^2 \cos(\omega t) \mbox{{}}$$

From this we see that the acceleration is always of opposite sign from the position: when the mass is above the zero position it is being accelerated downward, and when it is below its equilibrium position it is being accelerated upward. Also we see that the magnitude of the acceleration is largest when the velocity is $0$.

15.18   Definition (Acceleration due to gravity.) If a particle $\mbox{{\bf p}}$ moves near the surface of the earth, acted on by no forces except the force due to gravity, then $\mbox{{\bf p}}$ will move with a constant acceleration $-g$ which is independent of the mass of $\mbox{{\bf p}}$. The value of $g$ is

$$g =\frac{32 \mbox{ft./sec.}}{\mbox{sec.}} \mbox{(approx.)} \... ...g =\frac{9.8\mbox{meter/sec.}} {\mbox{sec.}} \mbox{(approx.)}.$$

We call $g$ the acceleration due to gravity. Actually, the value of $g$ varies slightly over the surface of the earth, so there is no exact value for $g$. The law just described applies in situations when air resistance and buoyancy can be neglected. It describes the motion of a falling rock well, but it does not describe a falling balloon.

Remark: When I solve applied problems, I will usually omit all units (e.g. feet or seconds) in my calculations, and will put them in only in the final answers.

15.19   Example. A juggler $J$ tosses a ball vertically upward from a height of $4$ feet above the ground with a speed of $16$ ft./sec. Let $h(t)$ denote the height of the ball above the ground at time $t$. We will set our clock so that $t=0$ corresponds to the time of the toss:

$$h(0) = 4; \hspace{4em} h'(0) = 16.$$

We will suppose that while the ball is in the air, its motion is described by a differentiable function of $t$. We assume that

$$h''(t) = -g = -32.$$

We know one function whose derivative is $-g$:

$$\mbox{ if } s(t) = -gt, \mbox{ then } s'(t) = -g.$$

By the antiderivative theorem it follows that there is a constant $v_0$ such that

$$h'(t) = s(t) + v_0 = -gt + v_0.$$

Moreover we can calculate $v_0$ as follows:

$$(16 =h'(0) = -g \cdot 0 + v_0) \mbox{$\Longrightarrow$}(v_0 = 16).$$

Thus

$$h'(t) = -gt + 16.$$

We know a function whose derivative is $-gt + 16$:

$$\mbox{ if } w(t) = -\frac{gt^2}{2} + 16t, \mbox{ then } w'(t) = -gt + 16.$$

Thus there is a constant $h_0$ such that

$$h(t) = w(t) + h_0 = -\frac{gt^2}{2} + 16t + h_0.$$

To find $h_0$ we set $t=0$:

$$( 4 = h(0) = -\frac{g \cdot0^2}{2} + 16\cdot 0 + h_0) \mbox{$\Longrightarrow$}(h_0 = 4).$$

Thus

$$h(t) = -\frac{gt^2}{2} + 16t + 4.$$

The ball will reach its maximum height when $h'(t) = 0$, i.e. when

$$t = \frac{16}{g} = \frac{16}{32} = \frac{1}{2}.$$

The maximum height reached by the ball is

$$h(\frac{1}{2}) = -\frac{1}{2} \cdot32 \cdot(\frac{1}{2})^2 + \frac{16}{2} + 4 = 8,$$

so the ball rises to a maximum height of $8$ feet above the ground.

15.20   Example (Conservation of energy.) Suppose that a particle $\mbox{{\bf p}}$ moves near the surface of the earth acted upon by no forces except the force of gravity. Let $v(t)$ and $h(t)$ denote respectively its height above the earth and its velocity at time $t$. Then

$$\frac{dv}{dt} = h''(t) = -g,$$

so

$$v \frac{dv}{dt} = -gv = -g \frac{dh}{dt}.$$

Now

$$v \frac{dv}{dt} = \frac{d}{dt}(\frac{1}{2}v^2),$$

so we have

$$\frac{d}{dt}(\frac{1}{2}v^2) = \frac{d}{dt}(-gh).$$

It follows that there is a constant $K$ such that

$$\frac{1}{2}v^2 = -gh + K,$$

or

$$\frac{1}{2}v^2 + gh = K.$$

If $m$ is the mass of the particle $\mbox{{\bf p}}$ then

$$\frac{1}{2}mv^2 + mgh = Km.$$ (15.21)

The quantity $\frac{1}{2}mv^2$ is called the kinetic energy of $\mbox{{\bf p}}$, and the quantity $mgh$ is called the potential energy of $\mbox{{\bf p}}$. Equation (15.21) states that as $\mbox{{\bf p}}$ moves, the sum of its potential energy end its kinetic energy remains constant.

15.22   Exercise. A A particle moves in a vertical line near the surface of the earth, acted upon by no forces except the force of gravity. At time $0$ it is at height $h_0$, and has velocity $v_0$. Derive the formula for the height of the particle at time $t > 0$.

15.23   Exercise. The acceleration due to gravity on the moon is approximately

$$g_m = .17g$$

where $g$ denotes the acceleration due to gravity on the earth. A juggler $J$ on the moon wants to toss a ball vertically upward so that it rises 4 feet above its starting height. With what velocity should the ball leave $J$'s hand?