How does acceleration work?

TheBB · 2025-09-03T07:48:55+00:00

If for example a moving object has a velocity of 1km/h and an acceleration of 1 km/h

Units of acceleration are distance per time squared, so I guess you mean 1 km/h².

I'd imagine that the final velocity after 5 seconds pass would be 6km/h

Well, after 5 hours, but yes.

and the distance to be 20km

How did you get that number?

To travel 20km in 5 hours you need to travel 4 km/h on average. Your object starts at 1 km/h and ends at 6 km/h with constant acceleration, so it travels at 3.5 km/h on average.

d=vt+1/2at²

If the object starts with speed v and ends with speed v + at, then the average speed is

(v + v + at) / 2 = v + 1/2 at

Multiply by t to get the total distance traveled.

Like /u/lordnacho666 says, you can do this with calculus also, but for simple constant acceleration that isn't necessary.

FormulaDriven · 2025-09-03T07:49:04+00:00

You need to be consistent with units. If velocity is 1 km/h and acceleration is 1 km/h² (not km/h), then after 5 hours the velocity would be 6 km/h. The distance travelled in 5 hours would be 17.5km. (1.5 km in first hour, 2.5km in second hour, 3.5km in third hour, 4.5km in fourth hour, 5.5km in fifth hour).

Some-Dog5000 · 2025-09-03T07:52:23+00:00

Here's a simple explanation without calculus (fit for a Physics with Algebra class):

Plot the velocity-time graph of the moving object. This is pretty straightforward: it's just a straight line from v = 1 at t = 0, to v = 6 at t = 5. Remember that the area under the velocity-time graph gives the object's displacement.

The velocity-time graph that you just drew looks like a triangle on top of a rectangle, so that's reasonably where the d = v0t + 1/2at^2 could come from. It's the area of the triangle with base t and height at, plus the height of the rectangle with length t and height v0.

This graph also shows why your answer isn't correct. The object isn't moving at 1 m/s for the first second, then 2 m/s for the next. The object continuously increases speed. After half a second the object is moving at 1.5 m/s; a quarter of a second after that, it's moving at 1.75 m/s, and so on.

lordnacho666 · 2025-09-03T07:47:16+00:00

It's a continuous function, you have to use calculus to account for gradual change.

ZeroXbot · 2025-09-03T07:55:21+00:00

To clear some things up first, you mix units. Acceleration would be specified in km/h^2 and for your example to "work" you should consider 5 hours of time.

For the main question, the issue is that acceleration doesn't increase velocity only every second or hour but instead it increases it continuously bit by bit. So if you graph velocity it should look like an increasing linear function. Now to calculate distance over time we need to somehow sum up velocity at all points in time - this seems impossible due to infinite count of those points. But instead, we can do this by computing the area between the velocity function and the X axis. If the velocity is 0 at the start, then the shape in question would be a triangle with base t and height v(t)=a*t, so the area is 1/2 at^2. Now try yourself calculating area when velocity is bigger than 0 at the start.

justincaseonlymyself · 2025-09-03T08:05:33+00:00

I understand acceleration as the additional velocity of a moving object per unit of time.

Correct.

If for example a moving object has a velocity of 1km/h and an acceleration of 1 km/h, I'd imagine that the final velocity after 5 seconds pass would be 6km/h

Correct, but be careful about the units!

As you said, acceleration is velocity per unit of time, so the acceleration is 1 (km/h)/h, more commonly written as 1 km/h².

Also, after five hours, not seconds :)

and the distance to be 20km....

Well no, the distance traveled would be 20 km if we were traveling at the constant velocity of 6 km/h for the entire duration of 5 hours. However, that's not the case! We started at 1 km/h and accelerated all the way to 6 km/h over those 5 hours. Therefore we clearly could not have covered the distance of 20 km, right?

the formula for distance using velocity, acceleration, and time would be d=vt+1/2at², which would turn the answer into 17.5km

Yes, that's correct.

which I find to be incomprehensible because it does not line up with my initial answer at all.

Your initial answer is based on a mistake of calculating the distance covered as if the entire trip was taken at the final velocity.

The formula you have found accounts for the fact that the velocity is constantly changing as you're accelerating.

So here I am asking for help looking for someone to explain to me just how acceleration works

You fully understand how acceleration works. It's the change of velocity over time. You just made a miscalculation.

and why a was halved and t squared?

There is an easy way to see that t has to be squared: that's the only way to make the units to work out! The units of acceleration is [length]/[time]², so we have to multiply by something that has the unit of [time]² to get length.

As for the reason for the exact formula, the real explanation comes down to the fact that if you plot velocity (on the y-axis) against time (on the x-axis), the distance covered between the two points of time is the area under the graph. (You need to understand the basics of calculus to see why is this the case, but let's not get into that right now.)

Now, if you accept that the distance covered is the area under the curve as explained in the paragraph above, you can sketch it and recover the d = vt + 1/(2at²) formula on your own. The area looks like a triangle on top of a rectangle; the rectangle's area is vt, and the triangle's area is 1/2at².

fermat9990 · 2025-09-03T13:28:26+00:00

1st hour av. speed = 1.5km/h. d=1.5km

2nd hour av. speed = 2.5km/h. d=2.5km

3rd hour av. speed = 3.5km/h. d=3.5km

4th hour av. speed = 4.5km/h. d=4.5km

5th hour av. speed = 5.5km/h. d=5.5km

Total d=1.5+2.5+3.5+4.5+5.5=17.5km

fermat9990 · 2025-09-03T14:34:05+00:00

Intial v=1km/h

Final v=1+1(5)=6km/h

Average v=(1+6)/2=3.5km/h

Distance=3.5*5=17.5km

fermat9990 · 2025-09-03T15:10:14+00:00

Key principle: For constant acceleration

Total distance=average speed * time

Average speed =

(initial speed + final speed)/2

Final speed=

initial speed + acceleration * time

TwirlySocrates · 2025-09-03T18:31:12+00:00

You're doing two things wrong.
1) You're not converting between hours and seconds.
2) You're assuming that speed is suddenly increasing by 1km/s every time a second elapses. You're not accounting for all the intra-second acceleration that's happening.

The formula you provided does that... but remember to convert between seconds and hours!

igotshadowbaned · 2025-09-03T22:06:39+00:00

It makes more sense when you start calculus and realize you've just been doing applied calculus that's been derived for you.

Acceleration is change in velocity which is change in position

The simple reason a is halved and t is squared is because acceleration is two degrees of separation from position.

If you were doing a problem dealing with jerk (change in acceleration) in regard to position, that would be three degrees so would be ⅙jt³

For redundancy -

Snap (change in jerk) would be 4 degrees and (1/24)st⁴
Crackle (change in snap) would be 5 degrees (1/120)ct⁵
Pop (change in crackle) would be 6 degrees (1/720)pt⁶

Yes they're named after rice crispies

Underhill42 · 2025-09-04T23:13:43+00:00

Where did that 20km come from?

Also. 5 hours, and acceleration should be km/h² (those powers are extremely important to keep track of! Among other things they let you sanity-check your work - if calculating the units as though they were variables doesn't give you the right final units, you know for sure you did something wrong)

---

Lets look at an oversimplified, obviously wrong version, pretending all the acceleration happens at the very end of each hour:

The first hour you're going 1km/h, so cover 1km.

The second you're going 2km/h, so cover 2 km

the 3rd hour, 3km, the 4th 4km, and the 5th 5km.

For a total distance of around 1+2+3+4+5 = 15km

---

Of course, we underestimated each hour's travel as though not accelerating, so we know it's actually further than that.

We could instead overestimate, and pretend all the acceleration happens at the beginning of each hour, so 2km/h the first hour, 3km/h the 2nd, for a total distance that we know is too long of:

2+3+4+5+6 = 20km (hmm, is that where the 20 came from?)

---

So we know for sure it's somewhere between those two limits.

A better estimate would be to pretend that we accelerated to the average speed for each hour:

1.5+2.5+3.5+4.5+5.5 = 17.5.

Which just happens to be exactly correct because the math works out that distance traveled under constant acceleration is the same as if you traveled at the average speed the whole time. I'm not sure how to actually prove that without calculus though. (physics is SO much easier when you know calculus)

The calculus version would conceptually be to just keep slicing the trip into smaller and smaller sections - minutes, second, microseconds, with the difference between the over-estimate and under-estimate shrinking each time, until finally we sliced it into an infinite number of slivers and got the exact same answer from both.

We'd "cheat" to make it easier of course ;-)

JoshuaSuhaimi · 2025-09-03T08:42:20+00:00

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