Newton's laws in three dimensional form

A 1200 kg car is being towed up an 18 degree incline by means of a rope attached to the rear of a truck. The rope makes an angle of 27 degrees with the incline. What is the greatest distance that the car can be towed in the first 7.5 s starting from rest if the rope has a breaking strength of 4600N?

If the answer is 11 m how would I get there?

~thanks.

we want to find the forces acting along the plane

the component of gravity acting down the plane is mg sin(theta), where theta = 18 deg
the component of force of the rope up the plane is F cos 27 where F is the force in the rope

if the max force in the rope is 4600N, then the max component of force up the plane is

4600cos27=4098.6N

newton's second law tells us that

4098.6 - mg sin 18 = ma

we can solve for a:

a=(4098.6N - 1200kg*9.8m/s/s sin 18)/1200 kg = 0.39m/s/s

now, recall from kinematics

dist = 1/2 at^2 = 1/2*0.39m/s/s*7.,5s^2 = 10.9 m...so close enough

draw a careful sketch. include direction of gravity, the weight

this is NOT three dimensional. it is two dimensional. everything is measured in the same flat plane vertical plane, the background, but the incline itself is a line on sketch

up down is gravity. and sketch the rest.
the distance is related to the maximum acceleration related to the maximum force along the ramp and the breaking strength of the rope.

it is complicated but the same old force balance and force and acceleration and distance problem in introductory physics