AS Maths question

A particle P moving with constant acceleration along a straight horizontal line ABC, where AC = 24m. Initially P is at A and is moving with speed 5ms^-1 in the direction AB. After 1.5s, the direction of motion of P is unchanged and P is at B with speed 9.5ms^-1.

a) Show that the speed of P at C is 13ms^-1. (4)
This part was fine.

The mass of P is 2kg. When P reaches C, an impulse of magnitude 30Ns is applied to P in the direction CB.

b) Find the velocity of P immediately after the impulse has been applied, stating clearly the direction of motion of P at this instant. (3)
Not so sure, can anyone help?

a)find the acceleration using v=u+at
9.5-4/1.5=3
then use V^2=u^2+2as to find v
sqrt((3*24*2)+25=13m/s
impulse=mv-mu
the speed of 13 is negative then remember that because it's in an opposite direction and it's u
30/2=v-(-13)
15=v+13
v=2m/s

in what direction is the particle travelling?