I started learning 1D kinematics and learnt definitions of constant velocity and constant acceleration. I thought that if a particle is changing its velocity uniformly then it has constant acceleration. From this I tried to plot x-t graph from the varying velocity. If a particle is moving with acceleration of $1 \,\text{m}/\text{s}^2$, then its velocity for first second would be $1 \,\text{m}/\text{s}$, $2 \,\text{m}/\text{s}$ for second second, $3 \,\text{m}/\text{s}$ for third and so on. Then I thought that since the particle is moving with $1 \,\text{m}/\text{s}$ at first second, it would have displaced $1 \,\text{m}$ and in second second, as its velocity was $2 \,\text{m}/\text{s}$ so it covered additional $2 \,\text{m}/\text{s}$ and so on an extra meter for each second than previous. In total, its displacement was $1 \,\text{m}+2 \,\text{m}+3 \,\text{m}$ that is $6 \,\text{m}$ of displacement. But actually the real displacement was $4.5 \,\text{m}$. How am I wrong?
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17$\begingroup$ The velocity is not changing in steps, but is changing continuously. Do you know calculus? $\endgroup$– Marius Ladegård MeyerCommented May 8 at 19:25
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2$\begingroup$ I don't see anything not focused about this question. $\endgroup$– Vincent ThackerCommented May 8 at 23:02
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1$\begingroup$ Use the formula $v=v_0 + a \Delta t$ to get a "feel" for what velocity is doing with respect to time. I suggest using various values, including non-integer values, for time when using this formula. $\endgroup$– David WhiteCommented May 9 at 15:36
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7$\begingroup$ "then its velocity for the first second would be 1 m/s": No! Its velocity would increase from 0 m/s to 1 m/s, only reaching 1 m/s after precisely 1 second (not before it). $\endgroup$– TonyKCommented May 9 at 17:47
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1$\begingroup$ Just to support the answers given, your approach is one of the very logical approaches we had to explore for literally thousands of years before calculus finally showed us how to get an answer that is consistent with physics for all possible acceleration curves. This is quite the fundamental question, actually! $\endgroup$– Cort AmmonCommented May 11 at 14:37
4 Answers
The (signed) area under the velocity-time graph, shaded in red, is the displacement.
What you think is happening:
What actually happens:
So the reason is that the velocity is changing continuously and not in steps. However, if you take smaller and smaller steps, your result approaches the true result, which is an integral.
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9$\begingroup$ Excellent pictures, and it would be useful to consider how, starting from the first picture where the velocity changes every second, you can go to a picture where the velocity changes every half-second, then to one where the velocity changes every quarter-second, and so on... until you arrive at the bottom picture as the limiting case. $\endgroup$– printfCommented May 9 at 13:27
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$\begingroup$ In a nutshell we used that summation>integration :) $\endgroup$ Commented May 9 at 18:29
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1$\begingroup$ @MathStackexchangeIsNotSoBad your inequality is true for the upper Riemann sum (the case here), but not in general. If we used the midpoint rule, the result would be exactly the same as integration. $\endgroup$– RuslanCommented May 10 at 16:05
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$\begingroup$ @Ruslan what is the midpoint rule? $\endgroup$ Commented May 10 at 16:26
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1$\begingroup$ @MathStackexchangeIsNotSoBad a method of quadrature. Just google for it, and you'll find tons of info about it. $\endgroup$– RuslanCommented May 10 at 18:51
To travel 1 meter in 1 second, an object needs to travel at an average speed of $1 \,\text{m}/\text{s}$. But an object accelerating from rest at $1 \,\text{m}/\text{s}^2$ does not have an average speed of $1 \,\text{m}/\text{s}$ over the first second, it has a top speed of $1 \,\text{m}/\text{s}$ - at all times before the first second is over, the object is moving slower than $1 \,\text{m}/\text{s}$. The object can't move slower than $1 \,\text{m}/\text{s}$ for the entire first second, yet find that it's moved $1 \,\text{m}$ in that second.
Over the first second, the object has an average speed of only $0.5 \,\text{m}/\text{s}$ (for constant linear acceleration, the average speed is simply the average of the initial and final speed). Over the second second, it has an average speed of $1.5 \,\text{m}/\text{s}$. Over the third second, it has an average speed of $2.5 \,\text{m}/\text{s}$. The overall average speed for the entire 3 seconds is $1.5 \,\text{m}/\text{s}$ (again the simple average of the final speed of $3 \,\text{m}/\text{s}$ and the initial speed of $0 \,\text{m}/\text{s}$) , yielding a total distance of $4.5 \,\text{m}$.
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$\begingroup$ You can also sum up the separate displacements in each second, remembering that within each second, the speed is the average between the beginning of the time interval and the end; for the first second, the average speed is (0+1)/2: The particle covers 0.5m. For the second second it's (1+2)/2 = 1.5 m/s (and hence m!), for the third it's (2+3)/2 = 2.5 m/s respectively meters, summed up 4.5 (also m/s, as you said, as well as meters if you add the partial distances). $\endgroup$ Commented May 11 at 18:43
There is nothing special about an interval of one second. $1 \ \mathrm{m/ s^{2}}$ does not literally mean a increase of $1 \ \mathrm{m/s}$ every second as in the first chart below. The real meaning is better approximated by an increase of $1/2 \ \mathrm{m/s}$ every $1/2$ second as in the second chart, or even better by an increase of $1/4 \ \mathrm{m/s}$ every $1/4$ second as in the third chart and so on.
A real object experiencing constant acceleration does not increase its velocity in discrete time steps, but continuously. Since distance is the product of velocity and time, the total distance travelled is the sum of the areas of the strips. The total distance travelled in the first chart is $6$. In the middle chart, the strip areas total to $5.25$. In the third chart the areas total to $4.5625$ and we are getting close to the exact solution already. If you keep doing this process of halving the intervals indefinitely, the total of the infinitesimal strips approaches the correct solution of $4.5$ in the limit. The calculus method of integration does this summation of the infinitesimal strips automatically for you, even if the acceleration is changing, so it is worth learning.
See it is just all about average velocity. The velocity increases from $0$ to $1 \ \mathrm{m/s}$ in $1$ second, that is what acceleration of $1 \ \mathrm{m/s^2}$ means. So in that one second, the velocity is not $1 \ \mathrm{m/s}$ at all points. On average it will be $0.5 \ \mathrm{m/s}$. In $0.1$ seconds, it will be $0.1 \ \mathrm{m/s}$. It gives the velocity at the instant and not over a time interval of $1$ second which I guess you are thinking it like.
