# A particle motion problem in the x-y plane with constant acceleration

• rudransh verma
In summary, at t=0, x=0, the object has a x-velocity of 7m/s and an x-acceleration of -9m/s². If the maximum x-coordinate reached is X, then the x-velocity is v=7i^ m/s when x=X, and the time at which x=X is t=0. The value of X is 2.7222...m.

#### rudransh verma

Gold Member
Homework Statement
From the origin, a particle starts at t=0 with a velocity v vector= 7i^ m/s and moves in xy plane with constant a= -9i^+3j^ m/s2. At the time the particle reaches the max x coordinate, what is its a)velocity and b)position vector?
Relevant Equations
a=(v-u)/t
s=ut+1/2at^2
I have attempted but I don’t get anywhere.

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Start by thinking about motion in the x-direction only (ignore the y-motion).

At t=0, x=0. Initial x-velocity is ##u_x## = 7m/s, and x-acceleration ##a_x## = -9m/s².

If the maximum x-coordinate reached is X then:
I) What is the x-velocity when x=X?
2) What is the time at which x=X?
3) What is the value of X?

If you can answer these, you can go onto the y-motion.

rudransh verma said:
Homework Statement:: From the origin, a particle starts at t=0 with a velocity v vector= 7i^ m/s and moves in xy plane with constant a= -9i^+3j^ m/s2. At the time the particle reaches the max x coordinate, what is its a)velocity and b)position vector?
Relevant Equations:: a=(v-u)/t
s=ut+1/2at^2

I have attempted but I don’t get anywhere.
I see that in your attempted solution, you are using vector, u, for the initial velocity rather than using v as in the given problem statement. That's fine as long you are consistent, although it's good practice for you to point that out.

I see that @Steve4Physics has just replied, so I will cut this short.

You might want to write your first "Relevant Equation" in a different form.
##\vec{v}(t)=\vec{u}+\vec{a}\ t ##

Also,
In general, vector division is an undefined operation.

Steve4Physics said:
If the maximum x-coordinate reached is X then:
I) What is the x-velocity when x=X?
2) What is the time at which x=X?
3) What is the value of X?
Now what?

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rudransh verma said:
Now what?
The working in your attachment is correct but you haven't spotted they key point.

If the maximum x-coordinate reached is X then:
1) What is the x-velocity when x=X?​

Hint: Imagine you are watching the process happening:
- an object is moving to the right with velocity 7m/s;
- a negative force (i.e. acting to the left) is applied, giving the object an acceleration to the left (i.e. a negative value of acceleration because F=ma).

How would you describe, in words, what happens to the object’s position and velocity? No calculations or equations are allowed!

Steve4Physics said:
It came to me and if I am right the velocity should be zero.

Steve4Physics
rudransh verma said:
It came to me and if I am right the velocity should be zero.
Yes! Another example is throwing a ball up. Initial velocity is positve, acceleration negative. The ball's speed decreases while rising, is zero at the highest point and increases as the ball comes back down.

Have a go at the rest of your question.

rudransh verma said:
@Steve4Physics But I have the answer and its not zero.
But you haven't worked out the y-velocity yet!

Steve4Physics said:
But you haven't worked out the y-velocity yet!
I have got the max x coordinate 2.7 m by v^2=u^2… But when I calculated the time t using s=ut+… I am getting two values .84 and .7 (quadratic).
Is the body moving in horizontal plane or vertical?

rudransh verma said:
I have got the max x coordinate 2.7 m by v^2=u^2…
The unrounded value is X=2.7222...m. Make sure you use the unrounded value in any calculations or you will introduce (possibly large) rounding errors.

rudransh verma said:
But when I calculated the time t using s=ut+… I am getting two values .84 and .7 (quadratic).
That's because you have (correctly) solved the incorrect equation, 4.5t² – 7t + 2.7 = 0. This has 2 solutions of approximately 0.848s and 0.706s.

You must solve (for example) 4.5t² – 7t + 2.7222222222222 = 0.

You will get two (almost) identical solutions. The two solutions will get closer to each other the more decimal places you use for X.

But a much better way to find t is to remember what you said in Post #7 and then use v = u + at.

rudransh verma said:
Is the body moving in horizontal plane or vertical?
We are not told if the plane is vertical, horizontal or tilted. It doesn’t matter. We are given the initial velocity and acceleration which are both in the xy plane. The body will only move in the xy plane. The orientation of the xy plane is therefore totally irrelevant. (It might be relevant if gravity were part of the problem, but it isn't.)

Steve4Physics said:
But a much better way to find t is to remember what you said in Post #7 and then use v = u + at.
I know u and a. What is t then ?

rudransh verma said:
I know u and a. What is t then ?
[Edited]
As noted in Post #12:
"...a much better way to find t is to remember what you said in Post #7 and then use v = u + at."

In the x-direction:
u = 7m/s (given in the question)
a = - 9m/s² (given in the question)
v = [the velocity when x=X, which you have already worked out in Post #7]
v = u + at

So you can work out t (the time to reach x=X).

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Steve4Physics said:
In the x-direction:
u = 7m/s (given in the question)
a = - 9m/s² (given in the question)
v = [the velocity when x=X, which you have already worked out in Post #7]
v = u + at

So you can work out t (the time to reach x=X)
I got it right .
But there is another way to do it which I don’t understand completely. I have attached the image. Explain!

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rudransh verma said:
I got it right .
But there is another way to do it which I don’t understand completely. I have attached the image. Explain!
I hate trying to respond to pictures. So I've converted some of it to text and Latex and included it in line.
the second picture said:
In order to solve this problem we will have to write down the radius vector of the particle as a function of time$$\vec{r}=\vec{v}t + \frac{1}{2}\vec{a}t^2$$
I prefer calling it a position vector since "radius" strikes me as relating to a circle.
the second picture said:
If we now insert the given values of acceleration and the velocity we obtain that$$\vec{r}=7t \hat{i}+\frac{1}{2}t^2(-9\hat{i}+3\hat{j})$$
After we regroup the terms we obtain that$$\vec{r}=(7t-\frac{9}{2}t^2)\hat{i} + \frac{3}{2}t^2\hat{j}$$
The maximum value of x coordinate we obtain after we use the mathematical definition of a maximum
The author is writing poorly. This is not the mathematical definition of a maximum. Instead, we are about to use one of the proven properties of a maximum of a function: A maximum must occur at an endpoint, at a discontinuity of some sort or at a point where the first derivative is zero. In this case we have no discontinuities or endpoints, so it will be at a point where the first derivative is zero. We will proceed to solve for the value of ##t## where the function derivative is zero.
the second picture said:
$$\frac{dx}{dt} = 0 = \frac{d(7t-\frac{9}{2}t^2)}{dt}$$ $$0 = 7 - 9t$$
From which we get that the x coordinate has its maximum at$$t=\frac{7}{9} \text{s}$$
Does this help at all?

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Steve4Physics
jbriggs444 said:
we are about to use one of the proven properties of a maximum of a function: A maximum must occur at an endpoint, at a discontinuity of some sort or at a point where the first derivative is zero. In this case we have no discontinuities or endpoints, so it will be at a point where the first derivative is zero. We will proceed to solve for the value of t where the function derivative is zero.
Can I think of it as at what point of time velocity become zero (dx/dt= v). Where x is the x component of r vector

rudransh verma said:
Can I think of it as at what point of time velocity become zero (dx/dt= v). Where x is the x component of r vector
Yes. The first derivative of position with respect to time is velocity. You are looking for the time value when the x component of velocity becomes zero.

rudransh verma said:
Can I think of it as at what point of time velocity become zero (dx/dt= v). Where x is the x component of r vector
You can also think of it by visualizing the motion this way. The particle starts at the origin moving in the positive x-direction at 7 m/s. Because the x-component of the acceleration is opposite to the velocity, the speed is decreasing which means that the particle will eventually stop moving in the positive x-direction and reverse. At this point in time, call it tmax, the particle will be at the maximum distance from the origin. What is the value of tmax? Well, if the particle starts at 7 m/s and loses 9 m/s for every second that goes by, how many seconds will have to elapse before the x-component of the velocity drops to zero? Note that after one second the particle has already reversed direction.

Steve4Physics
kuruman said:
You can also think of it by visualizing the motion this way. The particle starts at the origin moving in the positive x-direction at 7 m/s. Because the x-component of the acceleration is opposite to the velocity, the speed is decreasing which means that the particle will eventually stop moving in the positive x-direction and reverse. At this point in time, call it tmax, the particle will be at the maximum distance from the origin. What is the value of tmax? Well, if the particle starts at 7 m/s and loses 9 m/s for every second that goes by, how many seconds will have to elapse before the x-component of the velocity drops to zero? Note that after one second the particle has already reversed direction.
Makes sense. So body will stop before 1 sec because-9 is acceleration opposite to 7.
But I don’t think the body is moving in 1D and suddenly at .77 sec it changes its direction in y direction. There are y velocity components before .77 secs.

I was careful to say that the particle will stop moving in the x-direction. That doesn't mean that the particle will come to rest. The motion in the y-direction is separate and independent of the x-direction.

rudransh verma
kuruman said:
I was careful to say that the particle will stop moving in the x-direction. That doesn't mean that the particle will come to rest. The motion in the y-direction is separate and independent of the x-direction.
So if I am correct up to .77 sec it will move 2D and then at .77 it will return back and start in opposite direction

rudransh verma said:
So if I am correct up to .77 sec it will move 2D and then at .77 it will return back and start in opposite direction
The motion is always in 2D in the xy plane. At 0.77 s only the x-component of the velocity will change sign but not the y-component. Reversing direction in 2D means that both x and y components change their sign at the same time.

rudransh verma
jbriggs444 said:
ncluded it in line.
I prefer calling it a position vector since "radius" strikes me as relating to a circle.
I don’t understand How putting vectors in kinematic equation number 2 is right and getting position vector. Because these eqns are applicable in 1D motion. We use them when finding u or v or t or a in 1D.
These quantities are describing 1 D motion.
We can also see in projectile motion we divide it into two motions in x and y direction and put quantities as 1D motion.

rudransh verma said:
don’t understand How putting vectors in kinematic equation number 2 is right and getting position vector.
Which equation are you referring to? I don’t see any numbered equations in any of the posts or in your attachment.
Anyway, I don’t see why there would be an objection to writing a vector equation to describe the motion. You can write the position (or velocity, etc.) as one vector equation or two scalar ones.

haruspex said:
Which equation are you referring to? I don’t see any numbered equations in any of the posts or in your attachment.
Anyway, I don’t see why there would be an objection to writing a vector equation to describe the motion. You can write the position (or velocity, etc.) as one vector equation or two scalar ones.
I am referring to 2nd eqn of motion. I have used the eqn in describing 1D motion. How is it right to just use the vectors in this eqn because they don’t describe the 1D but 2D and 3D motion.

haruspex said:
You can write the position (or velocity, etc.) as one vector equation or two scalar ones.
Can I also write the position vector as two 1D 2nd eqn of motion ?

rudransh verma said:
Can I also write the position vector as two 1D 2nd eqn of motion ?
I do not know what you mean by "the second equation of motion". I am not aware of any standard ordering.
For a projectile, you could write e.g.
##y=u_yt-\frac 12gt^2, x=u_xt##
or in vectors ##\vec r=\vec ut+\frac 12\vec at^2, \vec a=-\hat yg##.

haruspex said:
For a projectile, you could write e.g.
y=uyt−12gt2,x=uxt
or in vectors r→=u→t+12a→t2,a→=−y^g.
Yes this eqn. Is there a way where we can derive/extract vector form from scalar form vice versa.

rudransh verma said:
Yes this eqn. Is there a way where we can derive/extract vector form from scalar form vice versa.
Sure.
To get the scalar eqns from the vector, take the dot product with, respectively, ##\hat x, \hat y##.
Conversely, multiply the scalar equns by respectively, ##\hat x, \hat y##, and add them to produce the vector equn.

@haruspex what path does the particle take in xy plane? Can you draw?

rudransh verma said:
@haruspex what path does the particle take in xy plane? Can you draw?
The acceleration is constant. What familiar path results from constant acceleration?

haruspex said:
The acceleration is constant. What familiar path results from constant acceleration?
Since the position vector is quadratic so it will be a parabolic path.

rudransh verma said:
What planet are you on?

SammyS, Steve4Physics and rudransh verma
haruspex said:
What planet are you on?
Not linear

rudransh verma said:
Since the position vector is quadratic so it will be a parabolic path.