Acky, wearing roller-skates, is standing beside his mailbox when his friend davey skates by at a constant speed of 4m/s. two seconds later, jacky skates after his friend and accelerates uniformly at 2m/s. how long does it take jacky to catch davey?

Explanation:

Let Jacky takes t seconds to catch up with his friend.

Distance traveled by him can be calculated as follows

s =ut + 1/2 a t²

u = 0 , a = 2m / s²

s = .5 x 2 t² = t².

Time during which Davey was in motion is t + 2

Distance covered by him

= 4 ( t + 2 )  

These two distances must be equal due to their meeting .

4(t +2  )= t²,

t² -4t -8 =0

t = 5.46 s

do you have a formula possibly