Kinematics problem (initial acceleration)

Jul 2016
17
0
A ball is rolling over a soccer field with constant velocity Vb > 0 at an angle of 45° to the goal line. It starts at the corner of the field (distance d from the middle of the goal line) at the time t0 = 0. At the same time a player starts heading towards the goal on a path perpendicular to the goal line. He starts from a position at distance 2d in front of the middle of the goal line) with ap(t) = a0 (1− (t/ß))

I've solved the problems but the left thing to do is just the initial acceleration. I use integration to find these solutions:

- Distance covered by the ball sb=Vb*t
- Time when ball is in front of the middle of the goal line tA = 2√d / Vb (with distance sA = d / cos45°)
- Velocity of player vp = a0 (t − (t^1/2ß))
- Distance covered by the player sp = a0 / 6*ß (3ßt^2−t^3)
- Velocity of the player when he reach the point (the point after the ball reach distance sA) vpA = a0*d / vb (√2 − (d/vb*ß)) *I'm inserting tA into vp, am I right?
- Initial acceleration of the player in order to reach the point (the point after the ball reach distance sA) at the same time as the ball a0 = ??

Any ideas would be very appreciated. Thanks.
 
Last edited:
Jun 2016
24
3
Everything looks good so far, although you have a typo in your velocity function above, should be
a0 (t − (t^2/2ß))
and not
a0 (t − (t^1/2ß))

I trust you had done that correctly, as your position function looks good, and it wouldn't have been with the typo. Your time at impact is right, as is your player's velocity at impact.

So now you need a plan to solve for a0. Here's a hint: how far will your player have traveled to reach the point of impact with the ball? You could produce a simple diagram to answer that question. Now that you know the player's position, can you see a way to solve for a0 at that moment in time?
 
Jul 2016
17
0
Okay, I've got the idea to put the tA into the Sp/t and this equation equals to Vb/2 (after I set the time of person and the ball are equals)
 
Jun 2016
24
3
You've got the right idea, in that you're going to use sp(tA), which will be the position of the player when striking the ball. Noticing that sp=0 when t=0, we can also say that this is the distance the player has traveled (as you had mentioned in your first post).

Now the trick will be to equate that to the correct value. This is where a diagram is very important, as it is in so many physics problems. I envision sitting on the sideline, with the goals on the left and right. The ball is in the lower left corner, and the player out in the middle of the field, a distance 2d in the x-direction from the left-hand goal line. The ball rolls out into the field at 45 degrees to the northeast. It travels a distance d in the y-direction according to the problem. How far did it travel in the x-direction? Do you see that the distance traveled by the player is now 2d minus this amount? Did your diagram help you see this?

As a physicist, it will help you a great deal to get into the habit of starting with a diagram. Details that can be difficult to envision in our heads become trivial when we see them in a picture with x- and y- axes.
 
Jul 2016
17
0
I've got the answer of the initial acceleration
a0 = Vb^2 / (d - (sqrt(2).d^2 / 3.ß.Vb))
as the result of 3 equations: setting tA = tP, vp = sp / t, and value of tA

Is there something wrong? By the way, thanks a lot for your help.