• songoku
In summary, any acceleration profile will work provided the delta v between interval start and interval end is 3 m/sec2.
songoku
Homework Statement
A particle moves along a straight line in such a way that its displacement during any given interval of 1 second is 3 meters larger than its displacement during the previous interval of 1 second. Which one of the following options is correct?
a. the particle moves with constant velocity 6 m/s
b. the particle moves with constant velocity 3 m/s
c. the particle moves with constant acceleration 3 m/s/s
d. the particle moves with constant acceleration 6 m/s/s
e. the particle moves with increasing acceleration over time
Relevant Equations
Kinematics
Let say the displacement of the first interval of 1 second is 10 m, so the displacement of the next 1 second interval will be 13 m, and the next after this one will be 16 m and so on.

Option (a) and (b) are wrong because in any interval of 1 second the displacement will be constant

I think option (c), (d) and (e) are also all wrong because the increment of displacement of each 1 second interval won't be constant 3 meters.

Am I wrong or actually no correct option? Thanks

I actually found that option c) is correct.

Define $$y(t)=v_0((t+1)-t)+\frac{1}{2}a((t+1)^2-t^2)$$, that is y(t) is the displacement between time t and t+1 (assuming motion with constant acceleration ##a## and initial velocity ##v_0##).
The problem tell us that ##y(t)=y(t-1)+3##. If you do the math you ll conclude that ##a=3##.

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songoku
songoku said:
I think option (c), (d) and (e) are also all wrong because the increment of displacement of each 1 second interval won't be constant 3 meters.

Am I wrong or actually no correct option? Thanks

If it travels ##3m## more every second, then it's traveling ##3m/s## faster every second, and that is pretty much the definition of an acceleration of ##3m/s^2##.

Is there any reason you chose not to do a calculation?

songoku
PeroK said:
If it travels ##3m## more every second, then it's traveling ##3m/s## faster every second, and that is pretty much the definition of an acceleration of ##3m/s^2##.

Is there any reason you chose not to do a calculation?
I think your reasoning is correct but for the average velocity and for the average acceleration. To prove that the acceleration is constant one has to do the math as i suggest in post #2.

PeroK
Delta2 said:
I actually found that option c) is correct.

Define $$y(t)=\frac{1}{2}a((t+1)^2-t^2)$$, that is y(t) is the displacement between time t and t+1 (assuming motion with constant acceleration a).
The problem tell us that ##y(t)=y(t-1)+3##. If you do the math you ll conclude that ##a=3##.

PeroK said:
If it travels ##3m## more every second, then it's traveling ##3m/s## faster every second, and that is pretty much the definition of an acceleration of ##3m/s^2##.

Is there any reason you chose not to do a calculation?

I did something like this:

Assume the object starts from rest and accelerates for ##3m/s^2##. Using kinematic formula:

##d = V_o ~ . ~ t + \frac{1}{2}at^2 = \frac{3}{2}t^2##

For ##t = 0 \rightarrow d = 0##

For ##t = 1 \rightarrow d = \frac{3}{2} m##

For ##t = 2 \rightarrow d = 12 m##

So I got that the object did not travel ##3 m## more each second

First of all for ##t=2## it is ##d=6m##. The problem tell us that it should be $$(\frac{3}{2}-0)+3=6-\frac{3}{2}$$ which holds.

songoku
songoku said:
I did something like this:

Assume the object starts from rest and accelerates for ##3m/s^2##. Using kinematic formula:

##d = V_o ~ . ~ t + \frac{1}{2}at^2 = \frac{3}{2}t^2##

For ##t = 0 \rightarrow d = 0##

For ##t = 1 \rightarrow d = \frac{3}{2} m##

For ##t = 2 \rightarrow d = 12 m##

So I got that the object did not travel ##3 m## more each second

If we assume ##a = 3m/s^2## and look at the velocity after each second we get:
$$t = 1s, v(1) = 3m/s; \ \ t = 2s, v(2) = 6m/s; \ \ t = 3s, v(3) = 9 m/s$$
Now we take the average velocity ##v_{avg}## during each second:
$$v_{avg}(0-1) = 1.5 m/s; \ \ v_{avg}(1-2) = 4.5 m/s; \ \ v_{avg}(2-3) = 7.5 m/s$$
And, we have the distance traveled every second:
$$d(0-1) = 1.5m; \ \ d(1-2) = 4.5m; \ \ d(2-3) = 7.5m$$
And we can see the distance traveled per second increases by ##3m## every second.

songoku
Delta2 said:
First of all for ##t=2## it is ##d=6m##. The problem tell us that it should be $$(\frac{3}{2}-0)+3=6-\frac{3}{2}$$ which holds.

PeroK said:
If we assume ##a = 3m/s^2## and look at the velocity after each second we get:
$$t = 1s, v(1) = 3m/s; \ \ t = 2s, v(2) = 6m/s; \ \ t = 3s, v(3) = 9 m/s$$
Now we take the average velocity ##v_{avg}## during each second:
$$v_{avg}(0-1) = 1.5 m/s; \ \ v_{avg}(1-2) = 4.5 m/s; \ \ v_{avg}(2-3) = 7.5 m/s$$
And, we have the distance traveled every second:
$$d(0-1) = 1.5m; \ \ d(1-2) = 4.5m; \ \ d(2-3) = 7.5m$$
And we can see the distance traveled per second increases by ##3m## every second.

Thank you very much for the help delta2 and perok

Delta2 and PeroK
Many non-constant acceleration profiles are possible.

The conditions of the problem are consistent with, for instance, a particle that is at rest for t=0 through t=0.5 and accelerates at 6 m/sec2 for t=0.5 through t=1.0, ending the interval at v=3 m/sec. The particle then repeats the acceleration profile forever, coasting for half a second and accelerating hard for another half-second.

Indeed, any acceleration profile will work, provided only that the delta v between interval start and interval end is 3 m/sec2 and that the profile is periodic with period 1 second.

One might imagine scenarios where velocity and acceleration are undefined everywhere. The problem statement only talks about "displacement". But this is a physics, not math, so I will not go there.

songoku
jbriggs444 said:
Many non-constant acceleration profiles are possible.

The problem does state:
A particle moves along a straight line in such a way that its displacement during any given interval of 1 second is 3 meters larger than its displacement during the previous interval of 1 second.

"during any given interval" rules out many interpretations.

songoku
hmmm27 said:
"during any given interval" rules out many interpretations.
Yes. It was tempting to think that it would rule out everything but a constant acceleration profile. Such is not the case. Though it does enforce periodicity.

hmmm27 and songoku
I don't think that any acceleration profile such that ##a(t)## is periodic with period 1sec and $$\int_t^{t+1} a(s)ds=3$$ (which simply means that the velocity is increasing by 3 units at the end of each second), would satisfy the condition of the displacement in any second being 3 units greater than the displacement in the previous second. It seems to work with piecewise constant accelerations but i don't think it will work with any acceleration curve.

songoku
Delta2 said:
I don't think that any acceleration profile such that ##a(t)## is periodic with period 1sec and $$\int_t^{t+1} a(s)ds=3$$ (which simply means that the velocity is increasing by 3 units at the end of each second), would satisfy the condition of the displacement in any second being 3 units greater than the displacement in the previous second. It seems to work with piecewise constant accelerations but i don't think it will work with any acceleration curve.
Why ever would it not work? If the acceleration is periodic (with period 1 second) and has an integral which is equal to 3 over any specific 1 second interval then it must have an integral which is equal to 3 over every 1 second interval.

It follows that the velocity at any time will be 3 m/s greater than the velocity one second prior.

It then follows that the displacement over any 1 second interval is 3 meters greater than the displacement in the 1 second prior.

songoku
jbriggs444 said:
It then follows that the displacement over any 1 second interval is 3 meters greater than the displacement in the 1 second prior.
i have serious doubts about this logical step. The displacement during 1sec depends on the velocity curve for that 1sec and not only on the start and end velocity.

songoku
Delta2 said:
i have serious doubts about this logical step. The displacement during 1sec depends on the velocity curve for that 1sec and not only on the start and end velocity.
$$\int_{T+1}^{T+2} v(t)\ dt = \int_T^{T+1}3+v(t)\ dt = \int_T^{T+1} 3\ dt + \int_T^{T+1}v(t)\ dt = 3 + \int_T^{T+1} v(t)\ dt$$
By the time we get to this step we've already reasoned that v(t) = 3 + v(t-1) for all t. Which is to say we've already constrained the entire velocity curve over the interval, not just its endpoints.

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songoku and Delta2
yes well i now realized that the velocity curve will also be kind of periodic (same waveform just shifted by +3 on the vertical axis, every 1 sec.), thanks for this.

songoku and jbriggs444

## 1. What is the definition of motion?

Motion is the change in position of an object over time relative to a reference point.

## 2. What is the difference between speed and velocity?

Speed is the rate at which an object covers distance, while velocity is the rate at which an object changes its position in a specific direction.

## 3. How is acceleration related to motion?

Acceleration is the rate at which an object's velocity changes over time. It is directly related to motion because it determines how quickly an object's speed or direction changes.

## 4. Can motion occur without a force acting on an object?

Yes, motion can occur without a force acting on an object. This is known as inertia, which is the tendency of an object to resist changes in its motion.

## 5. How do you represent motion on a graph?

Motion can be represented on a graph by plotting the position of an object on the y-axis and time on the x-axis. The slope of the line on the graph represents the object's velocity, and the line itself shows the object's direction of motion.

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