Two particles at different velocities collide. Find when and where they collide

I'm having a lot of trouble with these questions and any help would be appreciated.

Some examples:
1) Two hockey players are practising their shots. They are 90m apart and hit their balls on the ground directly towards each other. The first player hits his ball at 6ms-1 and the other hits hers at 4ms-1. Both balls decelerate at 0.1ms-2. Find the distance from the first player when the balls collide. Answer: 55m

2) A footballer kicks a ball directly towards a wall 10m away and walks after the ball in the same direction at a constant 2ms-1. The ball starts at 4ms-1 but decelerates at a constant rate of 0.5ms-2. When it hits the wall it rebounds the travel away from the wall at the same speed with which it hit the wall. Find the time after the initial kick when the ball returns to the footballer.
Answer: 4.17 seconds


Thanks in advance!

thank you so much!

thank you so much!

Can't say I fully agree with @Kvothe the Arcane on this one.

Yes, use $s=ut+\frac{1}{2}at^2$

These suvat equations assume s=0 when t=0.

For 1) you basically have two choices:

Either A)

Assigning the same positive direction for each player, and consider displacements relative to player1.

In which case for player1, $s_1=u_1t+\frac{1}{2}a_1t^2$ as before.

But for player2, since they are 90m away,

$s_2=90+u_2t+\frac{1}{2}a_2t^2$

And equate, $s_1=s_2$ etc. as before.

Or B)

Assign the positive direction towards the other player, i.e. it's different for each player.

Then $s_1,s_2$ are both given by the basic formula, and we look to solve $s_1+s_2=90$, i.e. they meet when the total distance travelled towards each other is 90m.

Can't say I fully agree with @Kvothe the Arcane on this one.

Yes, use $s=ut+\frac{1}{2}at^2$

These suvat equations assume s=0 when t=0.

For 1) you basically have two choices:

Either A)

Assigning the same positive direction for each player, and consider displacements relative to player1.

In which case for player1, $s_1=u_1t+\frac{1}{2}a_1t^2$ as before.

But for player2, since they are 90m away,

$s_2=90+u_2t+\frac{1}{2}a_2t^2$

And equate, $s_1=s_2$ etc. as before.

Or B)

Assign the positive direction towards the other player, i.e. it's different for each player.

Then $s_1,s_2$ are both given by the basic formula, and we look to solve $s_1+s_2=90$, i.e. they meet when the total distance travelled towards each other is 90m.

But still, if i follow this procedure, the answer comes exactly 4 seconds not 4.17 seconds