속도와 가속도

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.

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.