Kinematics problem (initial acceleration)

[rndaryam]

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.

 

[haruspex]

2√d

Did you mean that?

vp = a0 (t − (t^1/2ß))

Are you saying it is $a_0(t-(\frac {t^{\frac 12}}{\beta}))$?
I do not understand how you get that from ap(t) = a0 (1− (t/ß))

 

[jbriggs444]

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))

When posing these problems, it is good practice to define all of your variables. It is good form to use subscripts rather than mushing letters together and expecting the reader to understand the intended meaning.

Vb becomes V_b, the velocity of the ball on its 45 degree trajectory.
t0 becomes t₀, the starting time. It is unnecessary to specify that t₀ = 0, but it does not hurt either. [and eventually saves a fair bit of ink].
t becomes the elapsed time such that t=t₀ at the outset.
d is good as it stands. It is the width of the field and also the distance that the player stands away from the goal line at time t₀

That still leaves us guessing at what ap(t), a₀ and ß are supposed to mean.

Guess: ap(t) is the acceleration of the player at time t, more easily understood as a_p(t).
Guess: a₀ is a free parameter, the starting acceleration of the player.
Guess: This is multiplied by ${(1-\frac{t-t_0}{\beta}})$ to simulate a declining rate of acceleration (which may eventually go very negative).

It is also good form to specify what question is being asked. What is supposed to be calculated here?

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

So now s_b is the distance covered by the ball along its 45 degree trajectory. It is a function of time and should be expressed as s_b(t) if one is being picky.

- Time when ball is in front of the middle of the goal line tA = 2√d / Vb (with distance sA = d / cos45°)

So now we have t_A as the time when the ball is in front of the goal. But surely that should be directly proportional to d rather than directly proportional to the square root of d? Perhaps you meant to say $t_A = \frac{d\sqrt{2}}{V_b}$

- Velocity of player vp = a0 (t − (t^1/2ß))

This should be v_p(t). But I question the result of your integration. Surely you meant to write $a_0(t\ -\ \frac{t^2}{2\beta})$?

Note that you just broke your own case convention. You have V_b for the ball and v_p for the player. Not cool.

- Distance covered by the player sp = a0 / 6*ß (3ßt^2−t^3)

This seems right and recovers from the previous error. Repairing the formatting and re-distributing we have:

$s_p = a_0(\frac{t^2}{2} - \frac{t^3}{6\beta})$

- 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?

That should be v_p(t_A). Yes, you should be substituting t_A into the formula for v_p(t).

- 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 = ??

So you are looking for the required initial acceleration of the player in terms of $\beta$, d and V_b if the player is to arrive at the ball exactly as it crosses in front of the middle of the goal?

 

[rndaryam]

When posing these problems, it is good practice to define all of your variables. It is good form to use subscripts rather than mushing letters together and expecting the reader to understand the intended meaning.

Vb becomes V_b, the velocity of the ball on its 45 degree trajectory.
t0 becomes t₀, the starting time. It is unnecessary to specify that t₀ = 0, but it does not hurt either. [and eventually saves a fair bit of ink].
t becomes the elapsed time such that t=t₀ at the outset.
d is good as it stands. It is the width of the field and also the distance that the player stands away from the goal line at time t₀

That still leaves us guessing at what ap(t), a₀ and ß are supposed to mean.

Guess: ap(t) is the acceleration of the player at time t, more easily understood as a_p(t).
Guess: a₀ is a free parameter, the starting acceleration of the player.
Guess: This is multiplied by ${(1-\frac{t-t_0}{\beta}})$ to simulate a declining rate of acceleration (which may eventually go very negative).

It is also good form to specify what question is being asked. What is supposed to be calculated here?So now s_b is the distance covered by the ball along its 45 degree trajectory. It is a function of time and should be expressed as s_b(t) if one is being picky.So now we have t_A as the time when the ball is in front of the goal. But surely that should be directly proportional to d rather than directly proportional to the square root of d? Perhaps you meant to say $t_A = \frac{d\sqrt{2}}{V_b}$This should be v_p(t). But I question the result of your integration. Surely you meant to write $a_0(t\ -\ \frac{t^2}{2\beta})$?

Note that you just broke your own case convention. You have V_b for the ball and v_p for the player. Not cool.This seems right and recovers from the previous error. Repairing the formatting and re-distributing we have:

$s_p = a_0(\frac{t^2}{2} - \frac{t^3}{6\beta})$That should be v_p(t_A). Yes, you should be substituting t_A into the formula for v_p(t).So you are looking for the required initial acceleration of the player in terms of $\beta$, d and V_b if the player is to arrive at the ball exactly as it crosses in front of the middle of the goal?

Thank you so much for the hints, I already got all the answers. I'm so sorry because this is my first thread, will improve it for next time :)

Did you mean that?

Are you saying it is $a_0(t-(\frac {t^{\frac 12}}{\beta}))$?
I do not understand how you get that from ap(t) = a0 (1− (t/ß))

Also thanks for haruspex, I've already got the solution as jbriqqs444 gave me the hints to solve it :)