Suppose the distance $x$ varies with time as: $$x = 490t^2.$$ We have to calculate the velocity at $t = 10\ \mathrm s$. My question is that why can't we just put $t = 10$ in the equation $$x = 490t^2$$ which gives us total distance covered by the body and then divide it by 10 (since $t = 10\ \mathrm s$) which will give us the velocity, like this:- $$v~=~\frac{490 \times 10 \times 10}{10} ~=~ 4900\ \frac{\mathrm{m}}{\mathrm{s}}$$ Why we should use differentiation, like this: $$ \begin{array}{rl} x & = 490t^2 \\ \\ v & = \mathrm dx/\mathrm dt \\ & = \mathrm d(490t^2)/\mathrm dt \\ & = 490 \times 2 \times t \\ & = 490 \times 2 \times 10 \\ & = 9800\, \frac{\mathrm{m}}{\mathrm{s}} \end{array} $$ Which not only creates confusion but also gives different answer. Any help is highly appreciated.
7 Answers
Your question is legitimate and I don't understand why it got downvoted. The confusion arises in the difference between average and instantaneous velocity.
Consider this example: a car moves at 10 m/s for 5 seconds, then stops at a light for another five seconds. What is the velocity of the car after 7 seconds? According to your calculation, it would be $\frac{5 \,\textrm{s}\cdot10\,\textrm{m/s}}{7 \, \textrm{s}}\approx 7.14$ m/s, which is obviously wrong because the car is completely at rest after 7 seconds. What you just computed is the average velocity of the car during those 7 seconds.
Asking for the velocity of a body at a given point in time is equivalent to asking "how much will the position change after an infinitesimal amount of time?", which is, in non rigorous terms, like taking an infinitesimal amount of space $dx$ and dividing it by an infinitesimal amount of time $dt$ (this is not how derivatives actually mathematically are defined, but it works at an intuitive level). The average velocity during an infinitesimal amount of time becomes the instantaneous velocity and is computed using the derivative.
in our previous example we would obtain $0$, because at 7 seconds, and just before and just after 7 seconds, the car is at rest.
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One thing you should notice about your method is that you get a different result depending on what time range you're averaging over. You're averaging from time t = 0 to t = 10, but what's so special about t = 0?
If you do the same thing, but start from t = 5, you get:
$$v = \frac{490 \times 10 \times 10 - 490 \times 5 \times 5}{10 - 5} = 7350$$
Since the goal is to determine the instantaneous velocity at a particular time, the fact that the result depends on some other time that you include in the equation should be a strong indication that your result is not just about the desired time, it's about the range of time as a whole.
When you compute a derivative, you're calculating the limit of the result of this as the size of this time range gets smaller and smaller, which approaches the infinitessimal period that we call "instantaneous".
Another way to think about this intuitively is that the instantaneous velocity is what you would see if you had a speedometer and you looked at it at time t = 10. The speedometer reading at that time is not an average since you started the car, it's just that momentary velocity (this is a simplification, since the internal mechanism of the speedometer is necessarily averaging over a short period of time, but it gets the point across).
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You're calculating the difference quotient,$$ \frac{f\left(b\right)-f\left(a\right)}{b-a} \,,$$where $x\left(t\right)=490t^2$, $b=10$, and $a=0$ such that$$ \frac{f\left(b\right)-f\left(a\right)}{b-a}~=~ \frac{490 \times {10}^2-490 \times 0^2}{10-0}~=~ 4900. $$ The difference quotient converges on the derivative as the endpoints become infinitely close. Your calculation approach is pretty much like how computers perform parts of the finite difference method, if you make that interval smaller.
For example, let's use your method where $x_b=10+0.001$ and $x_a=10-0.001$; then, asking WolframAlpha to do this math for us, we've got$$ \frac{f\left(b\right)-f\left(a\right)}{b-a}~=~ \frac{490 \times {\left(10+0.001\right)}^2-490 \times {\left(10-0.001\right)}^2}{\left({10+0.001}\right)-\left({10+0.001}\right)}~\approx~ 9800. $$This is approximately the same value as we get from the analytical calculus approach.
That said, if an equation needs the instantaneous rate of change, $\frac{\mathrm{d}f\left(x\right)}{\mathrm{d}x}$, then that's what it needs. As you've noted, the difference quotient can be a very different number when the difference between $x_b$ and $x_a$ isn't negligibly small.
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Here you can see the position $x$ changing with respect to $t$ as $x=490t^2$.
You can see that the position changes faster on the right side. The velocity increases and the end velocity is obviously different from the initial velocity or the average one.
I hope the visualisation helps to reinforce what the other answers explain in words.
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As others have said, when you compute $$\frac{x(10)-x(0)}{10-0}$$ you are computing the average velocity over the 10 second interval from $t=0$ to $t=10$. Graphically, this corresponds to finding the slope of the secant line (line crossing the graph at 2 points) shown in the graph below:
The question, however, is asking for the instantaneous velocity at $t=10$, which corresponds graphically to the slope of the tangent line at $(10, 4900)$, as shown in the graph below:
If you plot both lines on the same graph you can see that they do not have the same slope; in fact the tangent line is exactly twice as steep as the secant line. This demonstrates visually that the instantaneous velocity is not the same as the average velocity.
Now here is a curious coincidence: the instantaneous velocity at $t=0$ is exactly $0$, the instantaneous velocity at $t=10$ is $9800$, so the average velocity on the interval $0 \le t \le 10$ happens to be equal to the average of the instantaneous velocities at the two endpoints of the interval! This does not happen in general -- in most cases, "average velocity over $[a,b]$" means something different from "average of the instantaneous velocities at $a$ and $b$" -- but it does always happen when the position function is quadratic (figuring out why this is true is a fun exercise).
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You are mixing up the difference between instantaneous velocity and average velocity. Your method looks at average velocity, which is the change in position divided by the time it takes to travel that distance. This does not tell us the velocity at t = 10 seconds though. In general, the object could be moving very fast, slowly, or even at rest at t = 10 but still have the same average velocity.
The time derivative of position gives us the instantaneous velocity at some time. So either method you described gives a velocity, they just do not describe the same things.
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If the speed is constant, average velocity and instantaneous velocity are the same, regardless which interval you use to calculate average velocity or which point you use to calculate instantaneous velocity.
In you example, the velocity is changing - you can tell because the distance is proportional to $t^2$, not to t - therefore, both average and instantaneous velocities will be different for different time intervals or different points in time, respectively.
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