Your question is:
"A car with a mass of 1350kg is tested by having it accelerated up a road with a slope of 1 in 5,increasing in speed from 2.5 to 12 m/s,covering a distance of 100m, agaisnt a frictional force of 750 N."
Now...to begin solving it, we need to first study the KINEMATICS of the situation. I.e. we need to mathematically describe the motion...and NOT YET care about what causes the motion.
The road distance is d
The initial speed is vi
The final speed is vf
We want the acceleration.
Use the standard kinematics formula:
vf^2 = vi^2 + 2*a*d
(derivation available on request...caution, you must understand calculus before attempting its derivation)
Solve for a:
a = (vf^2 - vi^2)/(2*d)
HANG ONTO THIS expression. We will use it later.
Later on, we desire the time interval. Let's solve for it.
a = (vf - vi)/t
t = (vf - vi)/a
t = 2*d*(vf - vi)/(vf^2 - vi^2)
Done with kinematics. Onto kinetics. What causes the motion?
Now, we need to assess each REAL FORCE acting on the car.
What real forces act on the car?
Weight (directly downward): equal to m*g...car's mass multiplied by Earth's gravitational field. Never equal to anything else.
Normal force (up and perpendicular to the road): Call it N.
What N does...is it will be as large as necessary to prevent the car from sinking through the asphalt.
Resistive force (down along the road): call it D for air drag.
This is a combination of reverse traction from rolling resistance and air drag...but for argument's sake, pretend it is all air drag. Also pretend that is independent of speed, which may not necessarily be true.
Traction force (up along the road): call it F
This is the force that a hypothetical strongman would need to push from behind in order to accomplish this. In REALITY of the car's operation, it is the road pushing forward on the car, because the wheels pushed backward on the road.