Very simple Accelerated motions

  • Thread starter dman5505
  • Start date
In summary, the problem involves two objects being released from the same point at different times and the goal is to find the time at which they have a vertical separation of L. By using the equations for position and simplifying the expression for the second object's position, the time at which L is achieved can be found to be t=L/(gt_0)+t_0/2.
  • #1
dman5505
1
0

Homework Statement



At t=0 an object is released from rest at the top of a tall building. At the time t_0 a second object is dropped from the same point. Ignoring air resistance, show that the time at which the objects have a vertical separation L is given by: t=L/(gt_0 )+t_0/2.

Homework Equations



L=v_0*t + (1/2)at^2

The Attempt at a Solution



With much fidgeting with the above equation, I decided that the distance must be equal to L=(1/2)gt^2-(1/2)g(t_0+t)^2. However when I simplified the equation, L=-L/(gt_0)-t_0/2 which is the equation I supposed to get multiplied by -1. Since the object is going down, maybe L is supposed to be -L?

More importantly, I was confused on how to represent the change in time for the second object and the first object.
 
Physics news on Phys.org
  • #2
Welcome to PF, dman.
I don't think your expression L=v_0*t + (1/2)at^2 can be correct because it doesn't include a variable t_0, which clearly affects L.
Why not start with an expression for the position of the first object as a function of time?
Then an expression for the position of the second. It will probably have a t - t_0 in it.
Finally, subtract the two to get L.
 
  • #3
Hi, I have the same problem, did you manage to solve it?
 
  • #4
Yes, it works out well. Show your work if you would like help with it.
 
  • #5



Your attempt at a solution is on the right track, but there are a few errors. Firstly, the equation you used, L=v_0*t + (1/2)at^2, is not applicable in this scenario because the objects are not thrown with an initial velocity. Instead, you should use the equation L=(1/2)gt^2, which represents the distance an object falls due to gravity in a given time t.

Next, you should consider the time at which the second object is released, t_0, as the starting point for your calculations. This means that t=0 for the second object, and t=t_0 for the first object.

Using the equation L=(1/2)gt^2, we can set up two equations, one for each object, as follows:

For the second object (released at t=0): L = (1/2)gt^2

For the first object (released at t=t_0): L = (1/2)g(t-t_0)^2

Now, we can set these two equations equal to each other to find the time at which the objects have a vertical separation L:

(1/2)gt^2 = (1/2)g(t-t_0)^2

Solving for t, we get t = L/(gt_0) + t_0, which is the same as the equation given in the homework statement.

To address your confusion about the change in time for the two objects, remember that the second object is released at t=0, so its time of motion is simply t. The first object, however, is released at t=t_0, so its time of motion is t-t_0.

I hope this helps clarify the problem and your approach to solving it. Keep up the good work!
 

1. What is accelerated motion?

Accelerated motion is the type of motion in which the speed or direction of an object changes over time.

2. How is acceleration calculated?

Acceleration is calculated by dividing the change in velocity by the time it takes for the change to occur. It is measured in meters per second squared (m/s²).

3. What is the difference between acceleration and velocity?

Acceleration is the rate of change of velocity, while velocity is the rate of change of position. In other words, velocity tells us how fast an object is moving and in what direction, while acceleration tells us how the velocity is changing.

4. Can an object have a constant velocity and still be accelerating?

Yes, an object can have a constant velocity and still be accelerating. This is because acceleration is not only dependent on the speed of an object, but also on its direction. So, if an object is moving in a circular path at a constant speed, it is considered to be accelerating because its direction is constantly changing.

5. What is the difference between positive and negative acceleration?

Positive acceleration occurs when an object's velocity is increasing, while negative acceleration (also known as deceleration) occurs when an object's velocity is decreasing. Positive acceleration is often associated with speeding up, while negative acceleration is associated with slowing down.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
21
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
710
Replies
8
Views
227
  • Introductory Physics Homework Help
Replies
14
Views
481
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
713
  • Introductory Physics Homework Help
Replies
26
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
286
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top