Help understanding this motion problem

[MihleDex]

Homework Statement

An object starts from rest at the origin at time t = −T and accelerates with constant acceleration a.
A second object starts from rest at the origin at time t = 0 and accelerates with the same a.
How far apart are they at time t? Explain the meaning of the two terms in your answer, both
physically in words and also with regard to the v vs t plots

Relevant Equations

Equations of motion.

From my understanding the the distance of the two at time t would be 0. Because at the statement it is said that both objects start at the origin. To me the two expressions mean the same if both objects start at the origin I don't see how it matters what value is assigned to t.

 

[etotheipi]

From my understanding the the distance of the two at time t would be 0.

At time $t$, for what time interval has the first object been accelerating? How far has it travelled? Also at time $t$, for what time interval has the second object been accelerating? How far has it travelled?

It mentions a $v$/$t$ graph, have you sketched this out already?

 

[MihleDex]

At time $t$, for what time interval has the first object been accelerating? How far has it travelled? Also at time $t$, for what time interval has the second object been accelerating? How far has it travelled?

It mentions a $v$/$t$ graph, have you sketched this out already?

Are your first question elaborating on what the homework question is asking?

 

[etotheipi]

Are your first question elaborating on what the homework question is asking?

I was asking you! The first object is traveling from time $-T$ to time $t$; so for how long has it travelled, and how far has it moved? Can you also work out how far the second object has moved, from time $0$ to time $t$?

Graphs look good, though it might be easier to do both lines on the same graph and make the size of the triangle a little larger! Just so that when you come to interpret your answer it's a bit easier.

 

[PeroK]

From my understanding the the distance of the two at time t would be 0. Because at the statement it is said that both objects start at the origin. To me the two expressions mean the same if both objects start at the origin I don't see how it matters what value is assigned to t.

Have you ever watched one of those big marathons where there are so many runners that they start at different times? The elite runners go first and the club runners start a bit later. They all start at the same starting line, but at different times. By the time the club runners start, the elite runners are already some way ahead.

They are definitely not all at the starting line at the same time. Your starting time matters.

 

[MihleDex]

I was asking you! The first object is traveling from time $-T$ to time $t$; so for how long has it travelled, and how far has it moved? Can you also work out how far the second object has moved, from time $0$ to time $t$?

The 1st object has traveled for t - (-T) = t + T
The 2nd object has traveled for t - 0 = t

The 1st object has traveled as distance of:
x = 1/2*a*(t+T)^2

The 2nd for:
x = 1/2*a*(t)^2

 

[etotheipi]

The 1st object has traveled for t - (-T) = t + T
The 2nd object has traveled for t - 0 = t

The 1st object has traveled as distance of:
x = 1/2*a*(t+T)^2

The 2nd for:
x = 1/2*a*(t)^2

That's right; what's the difference between those values (physically, the distance between them)? You'll want to expand the first expression.

 

[MihleDex]

That's right; what's the difference between those values (physically, the distance between them)? You'll want to expand the first expression.

The difference would be 2tT + T^2

 

[PeroK]

The difference would be 2tT + T^2

That's not a distance; that has units of time squared.

 

[MihleDex]

That's not a distance; that has units of time squared.

distance: atT +1/2(a*T^2)

Sorry I verified this time using Dimensional analysis.

 

[PeroK]

distance: atT +1/2(a*T^2)

Sorry I verified this time using Dimensional analysis.

Can you see another way to express that answer? Hint: what is the velocity of the first object at time $t = 0$?

This is what the question is asking you to do.

PS what do you notice about the second term? What does this represent physically?

 

[MihleDex]

Can you see another way to express that answer? Hint: what is the velocity of the first object at time $t = 0$?

This is what the question is asking you to do.

PS what do you notice about the second term? What does this represent physically?

Thanks a lot. The second term represents a distance and the velocity would be $1/2aT$ So i think another way to represent it would be $x = 1/2aT^2$

 

[etotheipi]

Have a look at your graph. Draw a vertical line at time $t$; you know the difference in displacement is the difference in area between the two lines. Can you see how these two terms

distance: atT +1/2(a*T^2)

match up to areas?

Thanks a lot. The second term represents a distance and the velocity would be $1/2aT$

Double check the half.

 

[PeroK]

Thanks a lot. The second term represents a distance and the velocity would be $1/2aT$ So i think another way to represent it would be $x = 1/2aT^2$

At $t = 0$, how far ahead is the first object, and what is its velocity?

 

[MihleDex]

Have a look at your graph. Draw a vertical line at time $t$; you know the difference in displacement is the difference in area between the two lines. Can you see how these two terms

match up to areas?Double check the half.

I combined the two graphs and drew the vertical line. I can see from the graph how the difference in displacement would be the difference in the area underneath the lines however I struggle to see the relationship to this expression: $atT +1/2(a*T^2)$

 

[MihleDex]

At $t = 0$, how far ahead is the first object, and what is its velocity?

I'd say it's $1/2aT^2$ afar and the velocity is $1/2aT$ is there something I'm missing?

 

[etotheipi]

I'd say it's $1/2aT^2$ afar and the velocity is $1/2aT$ is there something I'm missing?

Why are you putting the half in here for velocity? Final velocity after constant acceleration equals initial velocity (0) plus acceleration times time.

 

[MihleDex]

Why are you putting the half in here for velocity? Final velocity after constant acceleration equals initial velocity (0) plus acceleration times time.

My bad, I used the formula for average velocity instead of the one for instantaneous velocity.

 

[PeroK]

My bad, I used the formula for average velocity instead of the one for instantaneous velocity.

Let me explain part of the point of this question. If, at time $t =0$, one object is a point $x_0$, with velocity $v_0$, and a second object is at $x_0 = 0$ with zero velocity, and they both have the same subsequent acceleration $a$, then the common acceleration cancels out and the displacement between the objects is:
$$d(t) = x_0 + v_0t$$
In other words, the first object moves relative to the second with constant velocity.