Solved: 2D Kinematics Help | Average Speed, Decrease in Velocity

In summary, the conversation revolved around solving a problem involving a ball thrown from the ground and passing a window. The questions asked for the average speed of the ball as it passes the window, the magnitude of the decrease in velocity across the window, and the travel time between the top of the window and the ball's maximum height. The conversation included a discussion on the use of equations and solving for unknown variables, with various approaches and solutions presented. The final solution for the travel time was determined to be 0.48 seconds, using the formula t=-v_f/a.
  • #1
BitterSuites
38
0
[SOLVED] 2D Kinematics Help

Basically I want to make sure that I have worked this correctly and find out what I did wrong in Part 3, as my solution is not logical.

Problem Consider a ball thrown up from the ground. It passes a window in the time interval .251s. The distance across the window is 1.49m.

1) Find the average speed as the ball passes the window.
2) What is the magnitude of the decrease of the velocity across the window?
3) If the ball continues its path upward without obstruction, find the travel time between the top of the window and the ball's maximum height.

My Work

1)V = (Vo + Vf)/2 = (Xf + Xi)/t (This correspondence was the only thing I could come up with to solve this problem)

5.936 = (Vo + Vf)/2
2(5.936) - Vf = Vo

Vf = Vo +at
Vf = 2(5.936) - Vf + at
2Vf = 2(5.936) + at
Vf = 5.936 + (at/2)
Vf = 5.936 + (-9.8t/2)
Vf = 5.936 + ((-9.8 * .251)/2)
Vf = 4.706 m/s

2)Magnitude = absolute value of Vf - Vo

Vo = 2(5.936) - 4.706
Vo = 7.166

|Vf-Vo|=Magnitude
|4.706 - 7.166| = 2.46 m/s

3) x = (Vf^2 - Vo^2)/2a

x = (0 - 4.706^2)/(2 * -9.8)
x = 1.12992

dx = Vo*t + .5 at^2
dx = 4.706t + .5 * -9.8t^2
1.12992 = 4.706t - 4.9t^2
1.12992 = t(4.706 - 4.9t)
1.12992/t = 4.706 - 4.9t
(1.12992/t) + 4.9t = 4.706
1.12992 + 4.9t = 4.706t
1.12992 = -.194t
t = -5.82455

Obviously time shouldn't be negative. Did I do parts 1 & 2 correctly? Also, where did I go wrong with part 3?

Thanks in advance for any help.
 
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  • #2
BitterSuites said:
dx = 4.706t + .5 * -9.8t^2
1.12992 = 4.706t - 4.9t^2
1.12992 = t(4.706 - 4.9t)
1.12992/t = 4.706 - 4.9t
(1.12992/t) + 4.9t = 4.706
1.12992 + 4.9t = 4.706t

1.12992 = -.194t
t = -5.82455

I think you better double check the equations in bold. You're going to be solving a quadratic, and thus you should have two times, one negative and one positive.
 
  • #3
I'm being very dense for some reason. I have no idea what is wrong with it. :)
 
  • #4
How did you get rid of the "t" under "1.12992" ? If you multiplied both sides by "t", then how come the 4.9 didn't get any of that action?
 
  • #5
Wow. I can't believe I did that. I'll just blame the internet :D

Is there a better equation to use for "t"?

Also, do the previous two look okay?
 
  • #6
I don't believe there's a better equation, you'll just have to solve for the quadratic. I'll look over the other ones, but it may take me a minute.
 
  • #7
1) look again at the question. (vf+vo)/2 *is* the average speed across the window
2) you had the right form of the answer while you were trying to work out 1)
3) v=0 at max height - work from the top of the window to get there.

Regards,

Bill
 
Last edited:
  • #8
The average speed is given by [tex] v_{AVE} = \frac{distance}{time} [/tex]

Simply plug in the values.

Your answer for number 2 is correct.

Let me know what you get for number 3.
 
  • #9
For #3 I think you can use the equation:

[tex] v_2 = v_1 + at[/tex]

where:

[tex]v_2 = 0 m/s [/tex] (at max height velocity is 0 as antenna guy notes)
[tex] v_1 = 4.706 m/s [/tex] (I'm assuming the calculation for this is correct)
[tex] a = -9.8 m/s^2[/tex]

Then just solve for 't'.
 
  • #10
"1) look again at the question. (vf+vo)/2 *is* the average speed across the window"

The problem is that I'm not given V or Vf, so I couldn't use it to solve the problem. I only have Vo because I assume that it starts at rest.

After solving the quadratic, I got X1 = .48, X2 = .4804. I then threw it into a Quadratic Equation Solver and it agreed with that outcome.

The equation given by chocokat (which I didn't see until just now LOL) gives t = .4802. So should I be safe with .48s?

As for this equation:

"The average speed is given by Vave = d/t", would this work when the object has an unknown speed before it enters the field of view?

I'm sorry I'm being such a pain. I just have almost no confidence in my answers.
 
  • #11
BitterSuites said:
"1) look again at the question. (vf+vo)/2 *is* the average speed across the window"

The problem is that I'm not given V or Vf, so I couldn't use it to solve the problem. I only have Vo because I assume that it starts at rest.

Your assumption is incorrect, but your answer to (vf+vo)/2 is. v=dx/dt, and the window could be any height above the ground. vis:

v=dx/dt=1.49/0.251=5.96

BTW - dx=xf - xi (not " +")

The amount of work you put into answering 2) is commendable (although you did most of the work while trying to solve 1)). In the future, you might consider that:

vf-vo=a*t=-9.8*0.251=-2.46m/s


After solving the quadratic, I got X1 = .48, X2 = .4804. I then threw it into a Quadratic Equation Solver and it agreed with that outcome.

The equation given by chocokat (which I didn't see until just now LOL) gives t = .4802. So should I be safe with .48s?

I'm not sure you realize what you did to get your answer to part 3, so I'll try to describe what I see:
x = (Vf^2 - Vo^2)/2a
x = (0 - 4.706^2)/(2 * -9.8)
x = 1.12992

By substituting in 0 for Vf, I assumed you changed what Vf and Vo stood for. Vf is now v at max height, and you solved for the height above the window.

The quadratic approach you eventually used to resolve t from that point looks reasonable, but there is an easier way:

Let v_f be the velocity at the top of the window, and v_m be the velocity at max height.

v_m=0

v_m-v_f=a*t

-v_f=a*t

re-arranging:

t=-v_f/a

t=-(4.706)/(-9.8)=0.48s

As for this equation:

"The average speed is given by Vave = d/t", would this work when the object has an unknown speed before it enters the field of view?

Yes, it does work - and you can figure out the speed as the ball enters the window.

v=dx/dt=(xf-xo)/(tf-to)

vf=v+0.5*a*(tf-to)

vo=v-0.5*a*(tf-to)


I'm sorry I'm being such a pain. I just have almost no confidence in my answers.

Keep chugging - try to understand what the equations mean, and the confidence will come.

Regards,

Bill
 

Related to Solved: 2D Kinematics Help | Average Speed, Decrease in Velocity

1. What is 2D kinematics and why is it important?

2D kinematics is the study of motion in two dimensions, typically involving the use of vectors and equations to describe the position, velocity, and acceleration of an object. It is important because many real-world situations involve motion in more than one direction, and understanding 2D kinematics can help us analyze and predict the behavior of objects in these scenarios.

2. How do you calculate average speed in 2D kinematics?

To calculate average speed in 2D kinematics, you would first need to determine the total distance traveled by the object. This can be done by finding the magnitude of the displacement vector, which is the difference between the initial and final positions of the object. Then, you can divide the total distance by the total time taken to travel that distance to find the average speed.

3. What is the difference between average speed and average velocity?

Average speed is a measure of the total distance traveled divided by the total time taken, regardless of direction. Average velocity, on the other hand, takes into account both the magnitude and direction of an object's motion. It is calculated by dividing the total displacement by the total time taken.

4. How does a decrease in velocity affect an object's motion in 2D kinematics?

A decrease in velocity means that the object is slowing down, either due to a decrease in speed or a change in direction. This can affect an object's motion in 2D kinematics by changing its position, velocity, and acceleration over time. It can also impact the forces acting on the object and the resulting motion.

5. What are some real-world applications of 2D kinematics?

2D kinematics has many practical applications in fields such as engineering, physics, and sports. It is used in the design of structures and machines, the analysis of projectile motion, and the understanding of sports movements like throwing, jumping, and running. It is also important in fields such as navigation, robotics, and animation.

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