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A particle moves along a straight line with equation of motion $s=f(t)$, where $s$ is measured in meters and $t$ in seconds. Find the velocity and the speed when $t=5$, if $$f(t) = 100 + 50t - 4.9t^2$$

This problem seems incredibly simple to me, all I do is plug in $5$ for $t$ and I get $227.5$.

Simple, so I know that it went 227.5 meters in 5 seconds, simple enough. That gives me 45.5m/s except that is not the answer in the book. Where am I going wrong?

I am not sure what they mean by velocity, if they mean the average velocity or instaneous but since they didn't say I will use average since it is easier.

$100+50(a+h)-4.9(a+h)^2 - (100+50-4.9a^2)) / h $ I get $(-9.8ah + 4.9h^2 - 4.9a^2) /h$

Arturo Magidin
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  • Just because it's easier doesn't mean it's correct. I'm quite certain that the problem is asking for instantaneous velocity. – El'endia Starman Sep 11 '11 at 23:33
  • @Jordan: Since they said "velocity when $t=5$", they mean instantaneous velocity, not average velocity. To mean average velocity, they would need to tell you initial time and ending time, not just a single instant in time. – Arturo Magidin Sep 11 '11 at 23:36
  • I don't understand though, if s=f(t) then plugging in 5 into the equation gives me the distance, and 5 is the time. What is wrong with that? –  Sep 11 '11 at 23:36
  • @Jordan: Plugging $t=5$ into the equation gives you the position when $t=5$, not the distance traveled (the distance traveled is $f(5)-f(0)$). What is wrong is that you are computing average velocity over $0\leq t\leq 5$ (incorrectly), but they are asking you for instantaneous velocity at $t=5$. You are answering a different question. – Arturo Magidin Sep 11 '11 at 23:39
  • Ok, how do I know when to use average or instantaneous? Will average always give me multiple point? –  Sep 11 '11 at 23:44
  • @Jordan: if it is asked the speed, you should compute the instantanous speed. You only compute the average speed if it is explicitely asked to find the average speed. – Américo Tavares Sep 12 '11 at 00:16
  • And instantaneous is y2-y1/x2-x1 and average is the same as derivative right? –  Sep 12 '11 at 00:18
  • No, it's the other way round. The instantaneous speed is the derivative of $s$, i.e. $f'(t)$. The average speed between $t_1$ and $t_2$ is $\frac{f(t_2)-f(t_1)}{t_2-t_1}$. – Américo Tavares Sep 12 '11 at 00:38

3 Answers3

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The equation of the motion is

$$s=f(t) = 100 + 50t - 4.9t^2\text{ m}.$$

The velocity at the instant $t$ is by definition

$$\frac{ds}{dt}=f'(t) =50 - 4.9\times 2t\text{ m/s}.$$

At $t=5\text{ s}$

$$f'(5) =50 - 4.9\times 2\times 5=1\text{ m/s}.$$

Alternatively by the definition of the derivative, you get

$$\begin{aligned}f'(5)&=\lim_{h\rightarrow 0}\frac{f(5+h)-f(5)}{h}=\lim_{ h\rightarrow 0}\frac{ 227.5+h-4.9h^{2}-227.5}{h} \\ &=\lim_{h\rightarrow 0}\;1-4.9h=1\text{ m/s}.\end{aligned}$$

Detailed computation

$$\begin{eqnarray*} f(t) &=&100+50t-4.9t^{2} \\ f(5) &=&100+50\left( 5\right) -4.9\left( 5\right) ^{2}=227.\,5 \\ f(5+h) &=&100+50\left( 5+h\right) -4.9\left( 5+h\right) ^{2} \\ &=&227.5+h-4.9h^{2} \\ f(5+h)-f(5) &=&227.5+h-4.9h^{2}-227.\,5 \\ &=&h-4.9h^{2} \end{eqnarray*}$$

and $$\begin{equation*} \frac{f(5+h)-f(5)}{h}=\frac{h-4.9h^{2}}{h}=\frac{h(1-4.9h)}{h}=1-4.9h. \end{equation*}$$

Comment: The instantaneous velocity is the derivative of $s$, i.e. $f'(t)$. The average velocity between $t_1$ and $t_2$ is $\dfrac{f(t_2)−f(t_1)}{t_2−t_1}$.

Américo Tavares
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  • I don't understand why 4.9x5 is being used, shouldnt it be 4.9(5^2)? –  Sep 12 '11 at 00:12
  • @Jordan: I first computed by the rules of derivatives that you don't now. Because of that then I computed by the mathematical definition of derivative, as in your prior questions. I computed at $t=5$, i.e $f'(5)$. – Américo Tavares Sep 12 '11 at 00:21
  • To get the derivative part what happened? I am sure my math is wrong but did you just get f(5)+f(h)-f(5)? –  Sep 12 '11 at 00:30
  • @Jordan: The second computation I made (by the definition of derivative) I computed $\dfrac{f(5+h)-f(5)}{h}$ and let $h$ tend to $0$. – Américo Tavares Sep 12 '11 at 00:46
  • Is it clear now? – Américo Tavares Sep 12 '11 at 00:51
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When the problem says "find the velocity at $t=5$", they necessarily mean the instantaneous velocity (that is, the value of the derivative at $t=5$). This, because average velocities require two times: the initial time and the ending time.

If they were asking about the average velocity between $t=0$ and $t=5$, then the correct answer would be: $$\begin{align*} \text{avg velocity} &= \frac{\text{distance traveled}}{\text{time}} = \frac{\text{final position} - \text{initial position}}{5}\\ &= \frac{f(5)-f(0)}{5} = \frac{227.5 - 100}{5} = 25.5 \text{ m/s}. \end{align*}$$

But they are not asking this. They are asking for the instantaneous velocity when $t=5$. That is, they are asking for $f'(5)$. That means computing $$\begin{align*} f'(t) &= \lim_{h\to 0}\frac{f(5+h)-f(5)}{h}\\ &\strut\\ &=\lim_{h\to 0}\frac{\bigl( 100 + 50(5+h) - 4.9(5+h)^2\bigr) - \bigl(100 + 50(5) - 4.9(5^2)\bigr)}{h}\\ &\strut\\ &=\lim_{h\to 0}\frac{\bigl( 100 + 250 + 50h - 4.9(25 + 10h + h^2)\bigr) - \bigl(100 + 250 - 4.9(25)\bigr)}{h}\\ &\strut\\ &=\lim_{h\to 0}\frac{\bigl( 350 + 50h - 122.5 - 49h - 4.9h^2\bigr) - \bigl( 350 - 122.5\bigr)}{h}\\ &\strut\\ &=\lim_{h\to 0}\frac{227.5 + h - 4.9h^2 - 227.5}{h}\\ &\strut\\ &=\lim_{h\to 0}\frac{h - 4.9h^2}{h}. \end{align*}$$ Again, nothing but algebra so far. If we try evaluating at $h=0$, we get (as you might expect by now) $\frac{0}{0}$. But there is a very obvious factor of $h$ that can be cancelled, so we have: $$f'(5) = \lim_{h\to 0}\frac{h-4.9h^2}{h} = \lim_{h\to 0}\frac{1-4.9h}{1} = 1-4.9(0) = 1.$$

So the instantaneous velocity at $t=5$ is $1$. What are the units? The units of $f$ are meters, the units of $t$ are seconds. Since $f'(t)$ is computed as a fraction of (values of $f$)/(values of $t$), it is measured in (units of $f$)/(units of $t$) = meters/sec.

That is, the instantaneous velocity at $t=5$ is $1$ meter/sec.

What about the instantaneous speed? Velocity includes direction: positive velocity means movement left-to-right; negative velocity means movement right-to-left. Speed is just the size of the velocity. That is, the speed is the absolute value of the velocity (this is true for both average speed and for instantaneous speed). So the instantaneous speed at $t=5$ is $|f'(5)|$ meters per second, which in this case is $1$ meter per second, same as the velocity.

Arturo Magidin
  • 398,050
  • I thought "speed" and "velocity" had the same meaning. I corrected "speed" in my answer to "velocity". – Américo Tavares Sep 12 '11 at 01:21
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    @Américo: I believe the distinction is that in physics velocity is a vector, while speed is a scalar (namely, the norm of the velocity). – Arturo Magidin Sep 12 '11 at 03:22
  • In Portuguese we say "velocidade" (velocity) and "valor absoluto da velocidade" (absolute value of the velocity) or "magnitude da velocidade" (magnitude of the velocity). – Américo Tavares Sep 12 '11 at 09:19
  • @ArturoMagidin - Quick question: suppose the rate of change is $\frac{5}{3}$ m/s which we read as "five-thirds meters per second", is this equivalent to saying the rate of change is "five meters per three seconds"? – Taylor Rendon Jan 19 '21 at 21:19
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    @TaylorRendon: A rate of change of five meters every three seconds is exactly the same as a rate of 10 meters every six seconds is exactly the same as a rate of five thirds of a meter every second. – Arturo Magidin Jan 19 '21 at 21:33
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To obtain the instantaneous velocity, compute the derivative $f'(t) = d f(t) / dt$. The velocity at $t=5$ is $f'(5)$.

Jiri Kriz
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  • Or compute $f'(5)$... – Arturo Magidin Sep 11 '11 at 23:39
  • Is this the same as the instentaneous rate of change? Because for that I have (f(x2) - (f(x1) \ (x2 - x1) –  Sep 11 '11 at 23:42
  • @Jordan: yes, velocity is the instantaneous rate of position change. Because the position change is not linear in time, the velocity depends on the time. – Jiri Kriz Sep 11 '11 at 23:45
  • So I have to calculate another point? –  Sep 11 '11 at 23:46
  • How do I know what x1 and x2 are? I mean I have the original problem is that always x1? I am getting -3.9 for this using 5 and 6 for t. –  Sep 11 '11 at 23:51
  • No, just derive: f '(t) = 50 - 2 * 4.9 * t. Then evaluate f '(5) = 1 = velocity at 5s. – Jiri Kriz Sep 11 '11 at 23:53
  • @Jordan In average velocity problems, there are two times to consider. In instantaneous velocity problems, there is only one. What you need for this is the value of the derivative of $f$ when $t = 5$. That is, you want $$\lim_{t \rightarrow 5} \frac{f(t) - f(5)}{t-5}.$$ – Austin Mohr Sep 11 '11 at 23:56
  • @Austin Now I really don't know what to do. So I have f(5)-f(f)/5-5? –  Sep 11 '11 at 23:57
  • @Austin I see no examples, this book really is bad. –  Sep 12 '11 at 00:01
  • @Jordan See your earlier question: http://math.stackexchange.com/questions/63684/finding-fa-for-a-given-value. You want to do precisely this with $5$ as your value of $a$. – Austin Mohr Sep 12 '11 at 00:04
  • I am getting 100+50-4.9t^2-227.5 / (t-5) and I am stuck. –  Sep 12 '11 at 00:04
  • @Austin I thought I was doing instantaneous not average, that is average form. –  Sep 12 '11 at 00:05
  • @Austin But in my book the definition for instantaneous rate of change is different than average, I am asked for instantaneous but I am using average. –  Sep 12 '11 at 00:09
  • let us continue this discussion in chat – Austin Mohr Sep 12 '11 at 00:17