Kinematics problem -- Calculating acceleration from the trajectory

[Victoria_235]

Homework Statement

For each of the following cases, represent on the trajectory, the velocity and tangential and normal components of the acceleration at times t1 and t2 of a mobile that moves according to the graph of s (arc) versus t (time) on the left. The point O on the trajectory marks the origin
of arcs and the arrow indicates the direction of the rising arc. Indicate also when it is greater
the speed of the mobile, in t1 or t2.

Homework Equations

v= dS/dt

The Attempt at a Solution

I have tried it. But I don't understand what means when the graphic is drawn in the negative area.
I think there is many mistakes in my attempt.

 

[berkeman]

It is a bit difficult to understand the problem statement (at least for me). Is there a figure of the original trajectory plots S(t)? And if you only have motion in one dimension S(t), how can you have normal and tangential components to acceleration? Normal and tangential to what?

 

[Victoria_235]

In the figure on the right is represented the trajectory, and what the problem asks is to draw on it the components of the acceleration and the direction of the velocity. Also say if the speed is greater in t1 or in t2.

Yes, I upload the original plot.
Thank you in advance.

Thank you.

 

[I like Serena]

Hey Victoria!

Let's start with the speed on the leftmost graphs. Are you aware that the speed is the slope of the arc length graph? That is, a steep slope up or down is high speed, while a horizontal slope is zero speed.
Can we tell in each of the graphs where the slope steeper? In t1 or t2?

 

[haruspex]

But I don't understand what means when the graphic is drawn in the negative area.

Each of the trajectory charts has an origin and a direction marked. When the (s,t) graph goes negative it means it is at a point in the trajectory before the origin.
I see that in some cases it will traverse the same part of the trajectory more than once, so the "direction of the velocity" will be both ways there.

 

[Victoria_235]

Thank you!
I think I understand it. When the graph (S, t) is in the negative part, it is moving against the reference system, I understand that in the opposite direction to that indicated by the arrow in the drawing on the right.

 

[haruspex]

When the graph (S, t) is in the negative part, it is moving against the reference system, I understand that in the opposite direction to that indicated by the arrow in the drawing on the right.

That is not quite right - maybe it's not what you meant.
s is a position, not a speed. When it is in the negative, the position on the curve is before the origin. That is, to get there from the origin you would have to move against the arrow. But the particle may be moving either with the arrow or against it. That corresponds to the slope of the graph. If the slope is positive the particle is moving with the arrow.

 

[Victoria_235]

Ok, I think I have something...
It is ok?
Here is my attempt.

Thank you very much for your help :)

 

[haruspex]

Ok, I think I have something...
It is ok?
Here is my attempt.

Thank you very much for your help :)

I think you are having trouble separating conceptually the shape of the graph and the shape of the trajectory.

Look at t₁ in A. You have written (I think) ds/dt > 0. That would mean s is increasing, but at t₁ it is as far from O as it ever gets, so it cannot be increasing. What is the slope of the graph at t₁?

 

[Victoria_235]

In A, I thought speed is increasing from origin to t1, were it stops. From 0 to t1, I guess is going from O, in the trayectory to some point at t2.
In fact, I think I don't understand it... because how it is in my mind, in t2 should be again in O (in the trayectory draw).

 

[haruspex]

In A, I thought speed is increasing from origin to t

No, not the speed, the position.
In graph A, it starts at some point s₀ at time 0. This is where the graph crosses the vertical axis. s₀ is the height at which it crosses. In the trajectory, that means it is distance s₀ from O in the direction shown by the arrow.
We see from the graph that at time 0 the slope is positive, i.e. s is increasing. So it is moving away from O.
The slope decreases gradually, so it is moving more slowly.
At t₁ the slope is zero. The speed is the slope, always, so just for an instant it has stopped moving. It is now as far from O as it will get.

The question asks for the acceleration at this point. Acceleration is the rate of change of velocity. Velocity and acceleration are both vectors, so we have to worry about the directions.
What can you say about the direction of the velocity just before t₁?
What about just after t₁?

 

[Victoria_235]

Ok. I have tried again.
As you said, at first the particle should be moving with the arrow, because of the positive slope, then should reach some point at t1, which is the maximum distance to O. After t1, is moving against the arrow, until t1, which is in the origin O (where the arrow begins).

From t1 to t2 velocity is negative.

 

[haruspex]

Let's just get the positions at t₀ and t₁ right first.
I wrote that at t₀ it is

distance s₀ from O in the direction shown by the arrow

That is, start at O, move distance s₀ along the trajectory in the direction shown by the arrow to find where it is at t₀. You seem to have moved the other way.
Likewise, s₁ is further along in the same direction.

 

[Victoria_235]

Good evening,
Here is my last attempt.
Thank you very much for your help .

 

[haruspex]

Good evening,
Here is my last attempt.
Thank you very much for your help .

This is confusing. I just noticed that the diagrams of the question that you posted in #8 and #12 do not match any in post #3. According to #8 and #12, t₁ is at the top of the curve in the graph, but in #3 and #14 (for the sinusoidal graph) they are after the peak.

In your latest attempt, the solution for t₁ looks correct! You have velocity and tangential acceleration heading towards O and a normal acceleration towards the inside of the curve.
The t₂ answer is wrong, though. What is ds/dt there? What does that tell you about the normal (i.e. radial) acceleration?

In the second case, the inverted parabola graph, you have t₂ nailed but the same errors for t₁ as for t₂ in the above.

In the third case, the straight line graph, you have the location and velocity right for t₁. Since it is a straight line graph for s, what can you say about tangential acceleration?