Which statement is true? (Velocity and Acceleration of a Tennis Ball)

• paulimerci
In summary: Which statement is true about the tennis ball during this interval of time?(A) Its speed increases and it is changing its direction of travel.
paulimerci
Homework Statement
During an interval of time, a tennis ball is moved so that the angle between the velocity and the acceleration of the ball is kept at a constant 120º. Which statement is true about the tennis ball during this interval of time?
(A) Its speed increases and it is changing its direction of travel.
(B) Its speed decreases and it is changing its direction of travel.
(C) Its speed remains constant, but it is changing its direction of travel.
(D) Its speed remains constant and it is not changing its direction of travel.
Relevant Equations
No equations!
I understand that when a tennis ball is in motion, the velocity vectors and acceleration vectors are pointing in the same direction. When the ball slows down, it is decelerating and comes to a stop. In the above statement, I understand that from the given angle, both vectors are pointing in the opposite direction and so ball is decelerating. My question here is how do I know it changes its direction?

paulimerci said:
My question here is how do I know it changes its direction?
The direction of velocity changes due to acceleration which has the different direction of 120 degree, neither 0 nor 180 degree.

topsquark
anuttarasammyak said:
The direction of velocity changes due to acceleration which has the different direction of 120 degree, neither 0 nor 180 degree.
Thank you, I still don’t understand. Can you explain it in detail?

paulimerci said:
Homework Statement:: During an interval of time, a tennis ball is moved so that the angle between the velocity and the acceleration of the ball is kept at a constant 120º. Which statement is true about the tennis ball during this interval of time?
(A) Its speed increases and it is changing its direction of travel.
(B) Its speed decreases and it is changing its direction of travel.
(C) Its speed remains constant, but it is changing its direction of travel.
(D) Its speed remains constant and it is not changing its direction of travel.
Relevant Equations:: No equations!

I understand that when a tennis ball is in motion, the velocity vectors and acceleration vectors are pointing in the same direction.
That's totally untrue.
paulimerci said:
When the ball slows down, it is decelerating and comes to a stop. In the above statement, I understand that from the given angle, both vectors are pointing in the opposite direction and so ball is decelerating. My question here is how do I know it changes its direction?
Draw a diagram.

topsquark
PeroK said:
That's totally untrue.

Draw a diagram.
What was untrue in my statement? Just want to learn why my interpretation was wrong?

paulimerci said:
Thank you, I still don’t understand. Can you explain it in detail?
If acceleration has same (0 degree) direction with velocity, the direction of motion (i.e. direction of velocity) does not change and the speed increase. You drive a car with step on acceleration pedal.
If acceleration has opposite (180 degree) direction with velocity, the direction of motion (i.e. direction of velocity) does not change and the speed decrease. You drive a car with step on brake pedal.
In other cases acceleration has traverse component to velocity and it changes the direction of velocity. You drive a car with step on acceleration or brake pedal ( or even no steps on) and with turning steering wheel also.

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topsquark and paulimerci
paulimerci said:
What was untrue in my statement? Just want to learn why my interpretation was wrong?
For uniform motion in a circle, velocity and acceleration are perpendicular.

Or think about gravity acting on a tennis ball.

MatinSAR, topsquark and paulimerci
PeroK said:
For uniform motion in a circle, velocity and acceleration are perpendicular.

Or think about gravity acting on a tennis ball.
It’s a projectile motion problem isn’t it? When the ball is hit at an angle, the y component of velocity vector is opposite to acceleration due to gravity and so as it goes to max height the velocity becomes zero and the ball changes direction and falls to the ground. Am I correct?

paulimerci said:
Homework Statement:: During an interval of time, a tennis ball is moved so that the angle between the velocity and the acceleration of the ball is kept at a constant 120º. Which statement is true about the tennis ball during this interval of time?
(A) Its speed increases and it is changing its direction of travel.
(B) Its speed decreases and it is changing its direction of travel.
(C) Its speed remains constant, but it is changing its direction of travel.
(D) Its speed remains constant and it is not changing its direction of travel.
Relevant Equations:: No equations!

My question here is how do I know it changes its direction?

My first instinct in such problems would be to break up the acceleration vector into two perpendicular components such that one of the components is perpendicular to the velocity vector. Then simply make your conclusions to decide what happens and it will become an easy problem to solve. Keep in mind that if a vector's components are considered then you must ignore the original vector i.e the orginal vector has the same effect as it's components.

As others have suggested, drawing a diagram for the question is a must to understand how to proceed, else you will be stuck.

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paulimerci, DaveE, Steve4Physics and 1 other person
paulimerci said:
It’s a projectile motion problem isn’t it?
The problem in post #1 does not appear to be a projectile problem. The ball is moved by some unspecified force.
paulimerci said:
When the ball is hit at an angle
At an angle to what?
paulimerci said:
the y component of velocity vector is opposite to acceleration due to gravity
You originally wrote
paulimerci said:
the velocity vectors and acceleration vectors are pointing in the same direction
Now you appear to be limiting the discussion to their vertical components.
Either way, there is no particular relationship between the velocity and the acceleration. An object can be ascending or descending with downward or upward acceleration independently.
paulimerci said:
from the given angle, both vectors are pointing in the opposite direction
Not directly opposite, but you have right idea. Can you put it a little more accurately?
paulimerci said:
how do I know it changes its direction?
If the direction of travel does not change, what can you say about the direction of the acceleration? Think about a small interval of time, δt, and remember what the definition of acceleration is.

MatinSAR, PeroK, paulimerci and 1 other person
To @paulimerci :
Apply the rules shown below to the question, keeping in mind that velocity and acceleration are vectors.
1. If the acceleration of the object is not zero, its velocity has to change.
2. If the acceleration of the object is in the same direction as its velocity, its speed increases.
3. If the acceleration of the object is in the opposite direction to its velocity, its speed decreases.
4. If the acceleration of the object is perpendicular to its velocity, its speed stays the same.
It wouldn't be a bad idea to memorize these rules; they will serve you well.

DrClaude, MatinSAR, paulimerci and 2 others
haruspex said:
The problem in post #1 does not appear to be a projectile problem. The ball is moved by some unspecified force.

At an angle to what?

You originally wrote

Now you appear to be limiting the discussion to their vertical components.
Either way, there is no particular relationship between the velocity and the acceleration. An object can be ascending or descending with downward or upward acceleration independently.

Not directly opposite, but you have right idea. Can you put it a little more accurately?

If the direction of travel does not change, what can you say about the direction of the acceleration? Think about a small interval of time, δt, and remember what the definition of acceleration is.

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haruspex said:
The problem in post #1 does not appear to be a projectile problem. The ball is moved by some unspecified force.

At an angle to what?

You originally wrote

Now you appear to be limiting the discussion to their vertical components.
Either way, there is no particular relationship between the velocity and the acceleration. An object can be ascending or descending with downward or upward acceleration independently.

Not directly opposite, but you have right idea. Can you put it a little more accurately?

If the direction of travel does not change, what can you say about the direction of the acceleration? Think about a small interval of time, δt, and remember what the definition of acceleration is.
I think I had the wrong assumptions in mind.

paulimerci said:
I think I had the wrong assumptions in mind.
What happens during circular movement?
Could you have another acceleration vector that could be combined with the vector of centripetal acceleration in such a way that the resultant acceleration vector forms a constant angle with the velocity vector of the ball?

Lnewqban said:
What happens during circular movement?
Could you have another acceleration vector that could be combined with the vector of centripetal acceleration in such a way that the resultant acceleration vector forms a constant angle with the velocity vector of the ball?
I don’t understand. Why I have to consider circular motion here?

paulimerci said:
I don’t understand. Why I have to consider circular motion here?
You don’t have to, it is just a suggestion.

Could a purely accelerated rectilinear movement have acceleration and velocity vectors out of alignment?

Please, consider that forces are associated to accelerations (Newton’s law 2), but not to velocities.

Think of the forces you feel when traveling in a car.

paulimerci
The answer should be very easy if you look at the the parallel and perpendicular components of acceleration vector in the diagram you've drawn. Just apply points 3 and 4 to the parallel and perpendicular components respectively as mentioned in post#11 by @kuruman. Each component will exert it's effect independent of the other. There's no need to do any of the complex thinking like projectiles or centripetal acceleration.

DrClaude and kuruman
vcsharp2003 said:
The answer should be very easy if you look at the the parallel and perpendicular components of acceleration vector in the diagram you've drawn. Just apply points 3 and 4 to the parallel and perpendicular components respectively as mentioned in post#11 by @kuruman. Each component will exert it's effect independent of the other. There's no need to do any of the complex thinking like projectiles or centripetal acceleration.
Thank you! That makes much more sense.

vcsharp2003
paulimerci said:
Why I have to consider circular motion here?
In uniform circular motion, we see that the acceleration vector is always perpendicular to the velocity vector at all instants of time. It tells us what effect a perpendicular acceleration vector has on the velocity vector. And this effect is clearly mentioned in point 4 of post#11 by @kuruman. So, all your answers are in the points mentioned in post #11 and you need to simply apply the correct points to your question.

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paulimerci
Just out of curiosity, what is the path it is travelling if we assume a constant acceleration magnitude ##a##?

erobz said:
Just out of curiosity, what is the path it is travelling if we assume a constant acceleration magnitude ##a##?
You will also need to assume, say, planar motion.
It might be easier to start at one lower level of time differentiation, i.e. velocity making a constant angle to the position vector.

haruspex said:
You will also need to assume, say, planar motion.
It might be easier to start at one lower level of time differentiation, i.e. velocity making a constant angle to the position vector.
Thanks for the tip. I'll probably sleep on it and fail to produce it in the morning!

haruspex said:
You will also need to assume, say, planar motion.
If the acceleration is constant, the motion is sure to be planar.

nasu
jbriggs444 said:
If the acceleration is constant, the motion is sure to be planar.

Lnewqban said:

I was thinking something along the lines of this figure.

Obviously please look beyond the fact that my ##\vec{v'}## is not tangential to the path...it should be. My guess is some kind of a spiral trajectory.

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Lnewqban said:
I do not understand. You want to confine the motion to a horizontal plane? Or you want the acceleration to be confined to a horizontal plane? Yes. Either one implies the other.

jbriggs444 said:
If the acceleration is constant, the motion is sure to be planar.
My response was to
erobz said:
if we assume a constant acceleration magnitude a?
which I took to mean the magnitude was constant. That is confirmed by post #25.

I see it could be read as "assume a constant acceleration, magnitude a", but how can the acceleration be constant, nonzero, if at a constant angle to the velocity?

haruspex said:
My response was to

which I took to mean the magnitude was constant. That is confirmed by post #25.
Indeed, I took the passage to indicate constant acceleration rather than an acceleration of constant magnitude.

Yes, I see how an acceleration of constant magnitude at a constant angle to the velocity still has a "roll" degree of freedom if one does not confine the motion to a plane.

jbriggs444 said:
I do not understand. You want to confine the motion to a horizontal plane? Or you want the acceleration to be confined to a horizontal plane? Yes. Either one implies the other.
Sorry.
I just understand that for the condition of "the angle between the velocity and the acceleration of the ball is kept at a constant 120º" to be true, the vector of acceleration of gravity should not be coplanar with the movement, but perpendicular to the planar motion you have suggested.

Lnewqban said:
Sorry.
I just understand that for the condition of "the angle between the velocity and the acceleration of the ball is kept at a constant 120º" to be true, the vector of acceleration of gravity should not be coplanar with the movement, but perpendicular to the planar motion you have suggested.
We do not need to concern ourselves with forces here. It is purely a kinematic question. Somehow the acceleration is made to have this relationship to the velocity. We do not care how.

jbriggs444 and erobz
haruspex said:
We do not need to concern ourselves with forces here. It is purely a kinematic question. Somehow the acceleration is made to have this relationship to the velocity. We do not care how.
I see that applicable to D) only.
If the tennis ball is changing trajectory, forces and accelerations are acting upon it.

As I see it, for the angle between the constantly changing direction of the velocity and the acceleration vectors of the ball to be kept at a constant angle, the constant direction of the weight force disqualifies to be coplanar with the non-linear trajectory.

Lnewqban said:
I see that applicable to D) only.
No, it applies throughout.
Lnewqban said:
If the tennis ball is changing trajectory, forces and accelerations are acting upon it.
Sure, but it doesn’t matter what those forces are. All that matters is that as a result of them the ball moves as stated.
Lnewqban said:
, the constant direction of the weight force
Which maybe countered perfectly by a constant upward force as one of the forces contributing to the motion.
If I hold a ball and move it along the sides of a vertical square, its motion is described completely by its location as a function of time. Given that, we can find the velocity and acceleration functions. We do not need the forces to determine them.

I think I have a way (not saying it the best way) to the solution...in theory.

let ##y(x)## be the path the particle is on

##\vec{a}## has constant magnitude ##a##, but is fixed at angle ##\beta## relative to ## \vec{v}##.

The direction of ##\vec{v}## is given by the following relationship

$$\frac{dy}{dx} = \tan \theta \tag{1}$$

Accelerations:

$$a_x = a \cos ( \theta + \beta) = \frac{dv_x}{dt} \tag{2}$$

$$a_y = a \sin ( \theta + \beta ) = \frac{dv_y}{dt} \tag{3}$$

Using the Chain Rule and subbing (1):

$$\frac{dy}{dx} \frac{dx}{dt} = \tan \theta v_x = v_y \tag{4}$$

Sub (4) into (3):

$$\frac{d}{dt} ( \tan \theta v_x ) = a \sin ( \theta + \beta ) \tag{5}$$

Differentiate the LHS of (5):

$$v_x \sec^2 \theta \frac{d \theta}{dt} + \frac{dv_x}{dt} \tan \theta = a \sin ( \theta + \beta ) \tag{6}$$

Sub (2) into (6) for ##\frac{dv_x}{dt}## and rearrange:

$$\frac{d\theta}{dt} = \frac{a}{v_x \sec^2 \theta } (\sin ( \theta + \beta ) - \cos ( \theta + \beta) \tan \theta ) \tag{7}$$

Applying the Chain Rule to (7) to eliminate the parameter ##t##:

$$\frac{d \theta }{dx} = \frac{d\theta}{dt} v_x = \frac{a}{\sec^2 \theta } (\sin ( \theta + \beta ) - \cos ( \theta + \beta ) \tan \theta ) \tag{8}$$

In theory...solve (8) for ##\theta (x)##, and solve (1) to find ##y(x)##

I'm not brave enough to attempt an analytical approach. Perhaps it's just because trying cartesian coordinates it seems like an analytical pipe dream?

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I almost gave up too quick... That whole RHS reduces dramatically by applying the sum difference formulas:

$$\frac{d \theta }{dx} = a \sin \beta \cos \theta$$

I can get a little further:

$$\ln \left| \frac{ \sec \theta + \tan \theta }{ \sec \theta_o + \tan \theta_o} \right| = a \sin \beta ( x - x_o )$$

$$\sec \theta + \tan \theta = \left( \sec \theta_o + \tan \theta_o \right) e^{ a \sin \beta ( x - x_o ) }$$

Then let ##A## = RHS and multiply through by ## \cos \theta##:

$$1 + \sin \theta = A \cos \theta \implies 1 + \sqrt{1 - \cos^2 \theta} = A \cos \theta$$

That becomes:

$$\cos \theta \left( \left( A^2 + 1 \right)\cos \theta - 2 A \right) = 0$$

It should follow that ( ignoring solution ##\theta = 90^{\circ}##:

$$\cos \theta = \frac{2A}{A^2 + 1}$$

Then we can say that:

$$\tan \theta = \frac{ \sqrt{ (A^2 + 1)^2 - 4A^2}}{2A }$$

Now, we are back to where I was before. Theoretically, you could then plug that into (1) and integrate to get ##y(x)##.

$$y(x) = \int dy = \int \frac{\sqrt{ (A^2 + 1)^2 - 4A^2} }{2A} dx$$

Where A is given by:

$$A = \left( \sec \theta_o + \tan \theta_o \right) e^{ a \sin \beta ( x - x_o ) }$$

I thought it was kaput, until I noticed the difference of squares under the radical:

\begin{aligned} y(x) &= \int \frac{\sqrt{ ( (A^2 + 1) - 2A )( (A^2 + 1) + 2A) } }{2A} dx \\ \quad \\ &= \int \frac{\sqrt{ (A-1)^2 ( A+1)^2 }}{2A} dx \\ \quad \\ &= \int \frac{A-1}{2A} dx +\int \frac{A+1}{2A} dx \end{aligned}

Surely this is a complete blunder...I'm getting that:

$$\int \frac{A-1}{2A} dx +\int \frac{A+1}{2A} dx = \int dx = x-x_o$$

What...

$$y(x) = x - x_o$$

??

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haruspex said:
...
If I hold a ball and move it along the sides of a vertical square, its motion is described completely by its location as a function of time. Given that, we can find the velocity and acceleration functions. We do not need the forces to determine them.
Sorry @haruspex , I can’t understand your example.
Probably irrelevantly and wrongly, I have just tried to extend post #23 from @jbriggs444 by excluding the action of the unidirectional influence of g, as the problem in discussion gives the constrain of a rotating acceleration vector.

In my humble opinion, problems that may present vectors that change, rotate or move erratically without a natural reason are not helpful to students.
I hope that @paulimerci understands what the correct answer is and the natural or physical reason behind it.

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