Correct statement about this motion

[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

 

[Delta2]

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 $a=3$.

 

[PeroK]

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?

 

[Delta2]

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.

 

[songoku]

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 $a=3$.

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

 

[Delta2]

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]

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]

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.

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

 

[jbriggs444]

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.

 

[hmmm27]

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.

 

[jbriggs444]

"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.

 

[Delta2]

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.

 

[jbriggs444]

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.

 

[Delta2]

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.

 

[jbriggs444]

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.

 

[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.