속도와 가속도

False. The acceleration of the ball is given by $h''(t)=-9.8$. This is constant throughout its trajectory (and is due to gravity).

Remark: the velocity of the ball at $t=2$ is zero, since $h'(2)=-9.8(2)+19.6=0$, but the velocity is only zero for an instant. Since the velocity is changing, the acceleration is nonzero.

The acceleration is constant, which means the rate of change of the velocity is constant. So, since it took 10 seconds for the velocity to increase by 1 metre per second (from $1~\frac{\mathrm{m}}{\mathrm{s}}$ to $2~\frac{\mathrm{m}}{\mathrm{s}}$), then it always takes 10 seconds for the velocity to increase by 1 metre per second.

So, it takes 10 seconds to accelerate from $2~\frac{\mathrm{m}}{\mathrm{s}}$ to $3~\frac{\mathrm{m}}{\mathrm{s}}$. To accelerate from $3~\frac{\mathrm{m}}{\mathrm{s}}$ to $13~\frac{\mathrm{m}}{\mathrm{s}}$ (that is, to change its velocity by 10 metres per second), it takes $10\times 10=100$ seconds.

Let $s(t)$ be the distance your keys have travelled since they left your hand. The rate at which they are travelling, $s'(t)$, is decreasing by 0.25 metres per second. That is, $s''(t)=-0.25$. Therefore, $s'(t)=-0.25t+c$ for some constant $c$. Since $c=s'(0)=2$, we see $$s'(t)=2-0.25t$$ Then $s(t)$ has $2-0.25t$ as its derivative, so $s(t)=2t-\frac{1}{8}t^2+d$ for some constant $d$. At time $t=0$, the keys have not yet gone anywhere, so $0=s(0)=d$. Therefore, $$s(t)=2t-\frac{1}{8}t^2$$

The keys reach your friend when $s(t)=2$ and $t>0$. That is:

$$\begin{aligned} 2&=2t-\frac{1}{8}t^2\\ 0&=\frac{1}{8}t^2-2t+2\\ t&=8\pm4\sqrt{3} \end{aligned}$$

We need to figure out which of these values of $t$ is really the time when the keys reach your friend. The keys travel this way from $t=0$ to the time they reach your friend. (Then $s(t)$ no longer describes their motion.) So, we need to find the first value of $t$ that is positive with $s(t)=2$. Since $8-4\sqrt{3}>0$, this is the first time $s(t)=2$ and $t>0$. So, the keys take $$8-4\sqrt{3}\approx 1~\text{second}$$ to reach your friend.

Suppose the deceleration provided by the brakes is $d~\frac{\mathrm{km}}{\mathrm{hr}^2}$. Then if $v(t)$ is the velocity of the car, $v(t)=120-dt$ (at $t=0$, the velocity is 120, and it decreases by $d$ kph per hour). The car stops when $0=v(t)$, so $t=\frac{120}{d}$ hours.

Let $s(t)$ be the distance the car has travelled $t$ hours after applying the brakes. Then $s'(t)=v(t)$, so $s(t)=120t-\frac{d}{2}t^2+c$ for some constant $c$. Since $0=s(0)=c$, $$s(t)=120t-\frac{d}{2}t^2$$

The car needs to stop in 100 metres, which is $\frac{1}{10}$ kilometres. We already found that the stopping time is $t=\frac{120}{d}$. So:

$$\begin{aligned} \frac{1}{10}&=s\left(\frac{120}{d}\right)\\ \frac{1}{10}&=120\left(\frac{120}{d}\right)-\frac{d}{2}\left(\frac{120}{d}\right)^2\\ \\ \text{Multiplying both sides by $d$:} \\ \frac{d}{10}&=120^2-\frac{120^2}{2}\\ d&=5\cdot 120^2 \end{aligned}$$

So, the brakes need to apply 72 000 kph per hour of deceleration.

If the acceleration was constant, then it was $$\frac{17 500~\mathrm{mph}}{\frac{8.5}{60}~\mathrm{hr}} \approx 123 500 ~\frac{\mathrm{miles}}{\mathrm{hr}^2}$$

So, the velocity $t$ hours from liftoff is $$v(t)=123500t$$ Therefore, the position of the shuttle $t$ hours from liftoff (taking $s(0)=0$ to be its initial position) is $$s(t)=\frac{123500}{2}t^2=61750t^2$$ So, after $\frac{8.5}{60}$ hours, the shuttle has travelled $$s\left(\frac{8.5}{60}\right)=(61750)\left(\frac{8.5}{60}\right)^2\approx 1240 ~\mathrm{miles}$$ or a little less than 2000 kilometres.

The acceleration of a falling object due to gravity is $9.8$ metres per second squared. So, the object's velocity $t$ seconds after being dropped is $$v(t)=9.8t~\frac{\mathrm{m}}{\mathrm{s}}$$ We want $v(t)$ to be the speed of the peregrine's dive, so we should convert that to metres per second: $$325~\frac{\mathrm{km}}{\mathrm{hr}}\cdot \left(\frac{1000~\mathrm{m}}{1~\mathrm{km}}\right)\left(\frac{1~\mathrm{hr}}{3600~\mathrm{sec}}\right)=\frac{1625}{18}~\frac{\mathrm{m}}{\mathrm{s}}$$ The stone will reach this velocity when $$9.8t=\frac{1625}{18}\qquad\Rightarrow\qquad t=\frac{1625}{18(9.8)}$$

What is left to figure out is how far the stone will fall in this time. The position of the stone $s(t)$ has derivative $9.8t$, so $$s(t)=4.9t^2$$ if we take $s(0)=0$. So, if the stone falls for $\frac{1625}{18(9.8)}$ seconds, in that time it travels $$s\left(\frac{1625}{18(9.8)}\right)=4.9\left(\frac{1625}{18(9.8)}\right)^2\approx 416 ~\mathrm{m}$$ So, you would have to drop a stone from about 416 metres for it to fall as fast as the falcon.

Since gravity alone brings your cannon ball down, its acceleration is a constant $-9.8~\frac{\mathrm{m}}{\mathrm{s}^2}$. So, $v(t)=v_0-9.8t$ and thus its height is given by $s(t)=v_0t-4.9t^2$ (if we set $s(0)=0$).

We want to know what value of $v_0$ makes the maximum height 100 metres. The maximum height is reached when $v(t)=0$, which is at time $t=\frac{v_0}{9.8}$. So, we solve:

$$\begin{aligned} 100&=s\left(\frac{v_0}{9.8}\right)\\ 100&=v_0\left(\frac{v_0}{9.8}\right)-4.9\left(\frac{v_0}{9.8}\right)^2\\ 100&=\left(\frac{1}{9.8}-\frac{4.9}{9.8^2}\right)v_0^2\\ 100&=\frac{1}{2\cdot 9.8}v_0^2\\ v_0^2&=1960\\ v_0&=\sqrt{1960}\approx 44~\frac{\mathrm{m}}{\mathrm{s}} \end{aligned}$$

where we choose the positive square root because $v_0$ must be positive for the cannon ball to get off the ground.

The derivative of acceleration is constant, so the acceleration $a(t)$ has the form $mt+b$. We know $a(0)=-50~000$ and $a\left(\frac{3}{3600}\right)=-60~000$ (where we note that $3$ seconds is $\frac{3}{3600}$ hours). So, the slope of $a(t)$ is $\frac{-60~000+50~000}{\frac{3}{3600}}=-12~000~000$, which leads us to $$a(t)=-50~000-(12~000~000)t$$ where $t$ is measured in hours.

Since $v'(t)=a(t)=-50~000-(12~000~000)t$, we see $$v(t)=\frac{-12~ 000~000}{2}t^2-50~000t+c=-6~000~000t^2-50~000t+c$$ for some constant $c$. Since $120=v(0)=c$: $$v(t)=-6~000~000t^2-50~000t+120$$

Then after three seconds of braking,

$$\begin{aligned}v\left(\frac{3}{3600}\right) &=-6~000~000\left(\frac{3}{3600}\right)^2-50~000\left(\frac{3}{3600}\right)+120\\ &=-\frac{25}{6}-\frac{125}{3}+120\\ &\approx 74.2 \text{ kph} \end{aligned}$$

Remark: When acceleration is constant, the position function is a quadratic function, but we don't want you to get the idea that position functions are always quadratic functions–in the example you just did, it was the velocity function that was quadratic. Position, velocity, and acceleration functions don't have to be polynomial at all–it's only in this section, where we're dealing with the simplest cases, that they seem that way.

Different forces are acting on you (1) after you jump but before you land on the trampoline, and (2) while you are falling into the trampoline. In both instances, the acceleration is constant, so both height functions are quadratic, of the form $\frac{a}{2}t^2+vt+h$, where $a$ is the acceleration, $v$ is the velocity when $t=0$, and $h$ is the initial height.

• Let's consider (1) first, the time during your jump before your feet touch the trampoline. Let $t=0$ be the moment you jump, and let the rim of the trampoline be height $0$. Then, since your initial velocity was (positive) $1$ meter per second, your height is given by $$h_1(t)=\frac{-9.8}{2}t^2+t=t\left(\frac{-9.8}{2}t+1\right)$$ Notice that, because your acceleration is working against your positive velocity, it has a negative sign.
• We'll need to know your velocity when your feet first touch the trampoline on your fall. The time your feet first first touch the trampoline after your jump is precisely when $h_1(t)=0$ and $t>0$. That is, when $t=\frac{2}{9.8}$. Now, since $h'(t)=-9.8t+1$, $h'\left(\frac{2}{9.8}\right)=-9.8\left(\frac{2}{9.8}\right)+1=-1$. So, you are descending at a rate of 1 metre per second at the instant your feet touch the trampoline.

Remark: it is not only coincidence that this was your initial speed. Think about the symmetries of parabolas, and conservation of energy.

• Now we need to think about your height as the trampoline is slowing your fall. One thing to remember about our general equation $\frac{a}{2}t^2+vt+h$ is that $v$ is the velocity when $t=0$. But, you don't hit the trampoline at $t=0$, you hit it at $t=\frac{2}{9.8}$. In order to keep things simple, let's use a different time scale for this second part of your journey. Let's let $h_2(T)$ be your height at time $T$, from the moment your feet touch the trampoline skin ($T=0$) to the bottom of your fall. Now, we can use the fact that your initial velocity is $-1$ metres per second (negative, since your height is decreasing) and your acceleration is $4.9$ metres per second per second (positive, since your velocity is increasing from a negative number to zero): $$h_2(T)=\frac{4.9}{2}T^2-T$$ where still the height of the rim of the trampoline is taken to be zero.

Remark: if it seems very confusing that your free-falling acceleration is negative, while your acceleration in the trampoline is positive, remember that gravity is pushing you down, but the trampoline is pushing you up.

• How long were you falling in the trampoline? The equation $h_2(T)$ tells you your height only as long as the trampoline is slowing your fall. You reach the bottom of your fall when your velocity is zero. $$h_2'(T)=4.9T-1$$ so you reach the bottom of your fall at $T=\frac{1}{4.9}$. Be careful: this is $\frac{1}{4.9}$ seconds after you entered the trampoline, not after the peak of your fall, or after you jumped.
• The last piece of the puzzle is how long it took you to fall from the peak of your jump to the surface of the trampoline. We know the equation of your motion during that time: $h_1(t)=\frac{-9.8}{2}t^2+t$. You reached the peak when your velocity was zero: $$h_1'(t)=-9.8t+1 =0\qquad\Rightarrow\qquad t=\frac{1}{9.8}$$ So, you fell from your peak at $t=\frac{1}{9.8}$ and reached the level of the trampoline rim at $t=\frac{2}{1.98}$, which means the fall took $\frac{1}{9.8}$ seconds.

Remark: by the symmetry mentioned early, the time it took to fall from the peak of your jump to the surface of the trampoline is the same as one-half the time from the moment you jumped off the rim to the moment you're back on the surface of the trampoline.

• So, your time falling from the peak of your jump to its bottom was $\frac{1}{9.8}+\frac{1}{4.9}\approx 0.3$ seconds.

Let $v(t)$ be the velocity of the object. From the given information:

• $v(0)$ is some value, call it $v_0$,
• $v(1)=2v_0$ (since the speed doubled in the first second),
• $v(2)=2(2)v_0$ (since the speed doubled in the second second),
• $v(3)=2(2)(2)v_0$, and so on. So, for general $t$: $$v(t)=2^tv(0)$$ To find its acceleration, we simply differentiate. Recall $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\{2^x\}=2^x\log 2$, where $\log$ denotes logarithm base $e$. $$a(t)=2^tv_0\log 2$$ Remark: we can also write $a(t)=v(t)\log 2$. The acceleration doubles every second as well.