# Accelerated Motion

• zachcumer

#### zachcumer

Hey I have a quick question. If the length of an inclined plane, a little ball is rolling down, measures 25 meters, and its acceleration is 2 meters every second, the velocity at one second would be 2 meters a second. The velocity at 2 seconds, would be 4 meters per second, the velocity at 3 sec is 6 meters per second. the velocity at 4 sec would be 8 meters a second, and finally at five sec the velocity would reach 10 meters per second. Now the total distance the ball travels is 25 meters. They tell you that in two seconds the ball travels 4 meters. So the ball would travel 9 meters in 3 seconds, and 16 meters in the 4 second and 25 meters in the 5th second? Adding all those up you get a number way larger than 25 meters long...what am I doing wrong?

If the ball starts from rest and has a constant acceleration, the distance traveled is proportional to the square of the time since it started. So, say it travels 1 m in 1 second, then it will travel 4 meters in 2 seconds. Not 4 meters in the 2nd second. In the 2nd second, it will travel 4 - 1 = 3 meters.

Last edited:
What exactly is the question asking you?

IF the total length of the plane is 25m, then clearly after traveling 25m the acceleration of 2ms^-2 will no longer be the case.So after 5 seconds the acceleration will be different. Are you given the balls initially velocity or do we assume it to be zero? Would you mind to quote the full question.

I followed this formula: d = 1/2 * (A) * T^2

Here is the picture given: look at attached.

#### Attachments

• photo.jpg
36.9 KB · Views: 343
What exactly is the question asking you?

IF the total length of the plane is 25m, then clearly after traveling 25m the acceleration of 2ms^-2 will no longer be the case.So after 5 seconds the acceleration will be different. Are you given the balls initially velocity or do we assume it to be zero? Would you mind to quote the full question.

At rest. it says!

zachcumer, you seem to be interpreting d as the distance travelleed in the Tth second. It is not. It is the distance traveled in T seconds.

can you see my pic.

I followed this formula: d = 1/2 * (A) * T^2

Here is the picture given: look at attached.

First off, that assumes initial velocity to be zero, and that formal gives the total distance traveled from 0 to whatever t may be:

say you take you look at five seconds

Then the d in your equation is the total distance traveled from 0 seconds through to five seconds.

It is NOT the distance traveled during the fifth second!

EDIT: Ahh I see dx has beat me here.

This is exactly what my paper says: "An object starting from rest gains a speed v = at when it undergoes uniform acceleration. The distance it covers is d = 1/2 at^2. Uniform acceleration occurs for a ball rolling down an inclined plane. The plane below is titled so a ball picks up a speed of 2 meters every second; then it's acceleration a = 2 m/s^2. The positions of the ball are shown at 1 second intervals. Complete the six blank spaces for distance covered, and the four blank spaces for speeds."

This is exactly what my paper says: "An object starting from rest gains a speed v = at when it undergoes uniform acceleration. The distance it covers is d = 1/2 at^2. Uniform acceleration occurs for a ball rolling down an inclined plane. The plane below is titled so a ball picks up a speed of 2 meters every second; then it's acceleration a = 2 m/s^2. The positions of the ball are shown at 1 second intervals. Complete the six blank spaces for distance covered, and the four blank spaces for speeds."

To get the distance traveled in anyone second you find the d for t=x and t=x-1 and subtract the two:

Example
Your looking for the distance traveled during the 3rd second
Fill 3 and 4 in for t into your formula d=1/2at^2
and the difference in the two d's is the distance traveled in the 3rd second.

v = at. so a = 2 meters per second. time let's say equals one second. In one second the ball has traveled 2 meters. let's try that in two seconds. at = 2 * 2 = 4. 4 meters in two seconds...etc..right?

at is the velocity. Not the distance.

v = at. so a = 2 meters per second. time let's say equals one second. In one second the ball has traveled 2 meters. let's try that in two seconds. at = 2 * 2 = 4. 4 meters in two seconds...etc..right?

4 metres in the second second! In two seconds the ball has gone 6 metres! As dx pointed out that's a velocity [check the units]

You told us the formula d=1/2at^2 yet you used v=at?

Last edited:
d = 1/2 * a * t * t

distance in 5 seconds = d = 1/2 * 2 * 5 * 5...

Because the ball has rolled for 5 seconds.

Yes, that's right.

So...v= at ...d = 1/2 a * t * t...

V = at.

A = 2 meters in one second.

T = I dunno, 2 seconds.

2 * 2 = 4...

What are you trying to calculate? Is it the time taken to travel 25 meters? Also, a is 2 m/s in one second, not 2 m in one second. Remember, a is an acceleration.

I need to find the velocity of the ball at 2 second, 3 second 4 second, and 5 second. Also need to find distances at each second.

Look At The Picture!

You know that the velocity is increasing at the rate of 2 m/s per second. You also know that in the beginning, the velocity is zero. So, the velocity increases by 2 m/s every second. After 1 second it increases from 0 to 2 m/s. In the next second, it increases from 2 m/s to 4 m/s etc. You can find the distances at each second from the formula

$$s = \frac{1}{2} a t^2$$.

Thats all great and dandy however, look at my attached picture on the first page...I will attach it again..it gives you the length of 4 meters as the length in 2 seconds. Then it gives you a blank for you to fill in the length from the 1 second mark to the 2 second mark.

You know that the velocity is increasing at the rate of 2 m/s per second. You also know that in the beginning, the velocity is zero. So, the velocity increases by 2 m/s every second. After 1 second it increases from 0 to 2 m/s. In the next second, it increases from 2 m/s to 4 m/s etc. You can find the distances at each second from the formula

$$s = \frac{1}{2} a t^2$$.

That's the distance traveled after t seconds have passed! It's not the distance traveled at each second, the distance traveled at each second is the difference between say t=5 and t=4 plugged into that formula to get the distance traveled during the fourth second.

Here Look At This!

#### Attachments

• photo.jpg
36.9 KB · Views: 360
That's the distance traveled after t seconds have passed! It's not the distance traveled at each second, the distance traveled at each second is the difference between say t=5 and t=4 plugged into that formula to get the distance traveled during the fourth second.

Exactly. Thats what I said, and what I have said 2 times before in this thread.

Exactly. Thats what I said, and what I have said 2 times before in this thread.

Hate to be picky but that was what you said until:

You can find the distances at each second from the formula

$$s = \frac{1}{2} a t^2$$

Then you contradicted yourself :P but I know what you meant, I just was pointing it out in case zach followed it.

It's all cool I hope :)

*Like the sig by the way*

Sorry If I wasn't clear. Zach, calculate the distance that the object moves in 1 second from the time it started moving. Then calculate the distance that it moves in 2 seconds from the time it started moving. The difference of these values will give you the distance it moves in the 2nd second, i.e. the distance it moves in the time interval between the 1st second mark and the 2nd second mark.

can you guys see the pic?

can you guys see the pic?

No we can't, try hosting it on imageshack,photobucket etc etc

Not to worry though, we have a good idea as to what you were asked? Just carry out what dx has told you above.

My velocities from 1- 5 seconds are as follows: 2 m/s, 4 m/s, 6 m/s, 8 m/s, 10 m/s.
My Distances are as follows: ??

Take a look at the pic!

#### Attachments

• photo.jpg
27.1 KB · Views: 342
My velocities from 1- 5 seconds are as follows: 2 m/s, 4 m/s, 6 m/s, 8 m/s, 10 m/s.

Correct!

Sorry If I wasn't clear. Zach, calculate the distance that the object moves in 1 second from the time it started moving. Then calculate the distance that it moves in 2 seconds from the time it started moving. The difference of these values will give you the distance it moves in the 2nd second, i.e. the distance it moves in the time interval between the 1st second mark and the 2nd second mark.

Or alternatively use the velocitys for each second, find the distance traveled in that second, do this for all 5 seconds and then add em up.

Try doing it the way dx says though because, you wouldn't like to have to the summing for 30 or 40 seconds would you?

One thing: It says that the distance from zero to 2nd second is 4 meters. They also say the distance from zero to one second is 1 meter. So thinking logically I came up with 3 meters for the distance from 1 second to 2 second...Then it asks what the distance is from 2 second to 3 second(s)..then asks from 3 sec to 4 sec (length)..then from 4 sec to 5th sec...HELP!

As dx and me have been trying to tell you, to get the distance traveled in anyone second you merely have to find the total distance traveling in t seconds and the total distance traveled in t-1 seconds.

The distance traveled in the second t will be equal to the [Total distance traveled in (t) seconds]-[Total distance traveled in (t-1) seconds]

E.g take you times to be 1 second and 2 seconds.

In t seconds the total distance traveled from time t=t seconds to t=0 seconds equals
$$d=\frac{at^2}{2}=t^2$$

so:for t=1
$$d_1=\1^2=1$$

and for t=2
$$d_2=2^2=4$$

Thus for the second second, the distance traveled in was:

$$d_(t)-d_(t-1)=d_2-d_1= 4-1$$

EDIT: Dam Capslock :(

Last edited:
can't see what you wrote, can you post again

As dx and me have been trying to tell you, to get the distance traveled in anyone second you merely have to find the total distance traveling in t seconds and the total distance traveled in t-1 seconds.

The distance traveled in the second t will be equal to the [Total distance traveled in (t) seconds]-[Total distance traveled in (t-1) seconds]

E.g take you times to be 1 second and 2 seconds.

In t seconds the total distance traveled from time t=t seconds to t=0 seconds equals
$$d=\frac{at^2}{2}=t^2$$

so:for t=1
$$d_1=1^2=1$$

and for t=2
$$d_2=2^2=4$$

Thus for the second second, the distance traveled was:

$$d_{(t)}-d_{(t-1)}=d_2-d_1= 4-1$$

EDIT: Dam Capslock :(

Fixed, sorry bout that :)

Last edited:
So its 3 meters~!~ YA!