I neer help in a mechanics question so urgently !

UVATS equations, don't forget to convert km to m and hours to seconds
I believe there will be simultaneous equations,

the journey has two parts and a is constant, thus you need to make two equations.
the first equation at p to q;
u1 = ? v1=? a = a t= 1 s= 2.4. (1)
the second equation at q to r;
u2= ? v2= ? a =a t=1.5 s = 11.5 ( 2)
at first they look different however, v1 is equal to u2 v1=u2, this is because the final velocity at q after travelling p to q will be the initial velocity from q to r ( the starting point is the ending point)
with this fact use the uvat equation that removes U in equation (1) to find v1, then use the equation that removes V in (2)
this will be s=vt-1/2at^2 and s=ut+at^2
this should leave you with two equations; after rearranging
v1= 2.4 - 1/2a (3)
u2 = 23/3 - 0.75a (4)
v1 = u2 from earlier therefore, 2.4 - 1/2a = 23/3 - 0.75a
rearrange for a = 4.21333333333333333 km h^-2 recurring

In an orienteering competition, a competitor moves in a straight line past three check points P,Q and R. Where PQ =2.4 Km and QR = 11.5 KM. The competitor is modelled as a particle moving with constant acceleration. She takes 1 hour to travel from P to Q and 1.5 hours to travel from Q to r find;
The acceleration of the competitor

When you have a question with a few unknowns safely assume you'll be using simultaneous equations, now first consider the motion from P-Q.
S=2.4, U=x, a=Y, t=1
Now using S=ut+1/2at^2
2.4=x +1/2y (1)
Since you want your initial velocity to be the same consider the motion from P-R
Therefore,
S=13.9, U=x, a=y, t= 2.5
Now using S=ut+1/2at^2
13.9=2.5x+3.125y (2)
Rearrange (1) to find X
2.4-1/2y=x
Now sub into (2)
13.9=2.5(2.4-1/2y)+3.125y
13.9= 6-1.25y+3.125y
7.9= 1.875y
Y= 4.21kmhr^-2

the journey has two parts and a is constant, thus you need to make two equations.
the first equation at p to q;
u1 = ? v1=? a = a t= 1 s= 2.4. (1)
the second equation at q to r;
u2= ? v2= ? a =a t=1.5 s = 11.5 ( 2)
at first they look different however, v1 is equal to u2 v1=u2, this is because the final velocity at q after travelling p to q will be the initial velocity from q to r ( the starting point is the ending point)
with this fact use the uvat equation that removes U in equation (1) to find v1, then use the equation that removes V in (2)
this will be s=vt-1/2at^2 and s=ut+at^2
this should leave you with two equations; after rearranging
v1= 2.4 - 1/2a (3)
u2 = 23/3 - 0.75a (4)
v1 = u2 from earlier therefore, 2.4 - 1/2a = 23/3 - 0.75a
rearrange for a = 4.21333333333333333 km h^-2 recurring

the journey has two parts and a is constant, thus you need to make two equations.
the first equation at p to q;
u1 = ? v1=? a = a t= 1 s= 2.4. (1)
the second equation at q to r;
u2= ? v2= ? a =a t=1.5 s = 11.5 ( 2)
at first they look different however, v1 is equal to u2 v1=u2, this is because the final velocity at q after travelling p to q will be the initial velocity from q to r ( the starting point is the ending point)
with this fact use the uvat equation that removes U in equation (1) to find v1, then use the equation that removes V in (2)
this will be s=vt-1/2at^2 and s=ut+at^2
this should leave you with two equations; after rearranging
v1= 2.4 - 1/2a (3)
u2 = 23/3 - 0.75a (4)
v1 = u2 from earlier therefore, 2.4 - 1/2a = 23/3 - 0.75a
rearrange for a = 4.21333333333333333 km h^-2 recurring

Thank you so so so much for this i cant stress how thankful i am. God bless u !

Can sum1 rearrange it bc I'm getting 5.26666=.25a which is wrong