Step-by-step explanation:
I know you've already done parts a and b, but I'll show the work for that before I do c.
Draw two free body diagrams, one for the car and one for the trailer. The car pulls the trailer forward with a tension force T, so the trailer pulls backward on the car with an equal and opposite force T.
The car also has a 1200 N forward force from the engine, and a 200 N backwards force from resistance.
The trailer has a backwards resistance force of 100 N.
Sum of forces on the car:
∑F = ma
1200 − 200 − T = 900a
1000 − T = 900a
Sum of forces on the trailer:
∑F = ma
T − 100 = 300a
To solve the system of equations, first add the equations together.
1000 − 100 = 1200a
900 = 1200a
a = 0.75 m/s²
Plug back into either equation to find the tension force:
T = 325 N
Now for part c, draw new free body diagrams for the car and trailer. This time, the car is pushing back on the trailer to slow it down. So the trailer is pushing forward on the car with an equal and opposite force. The magnitude of that tension force is given to be 100 N.
The car also has a backwards 200 N force from resistance, and a backwards brake force F.
The trailer has a backwards 100 N force from resistance.
Sum of forces on the car:
∑F = ma
100 − 200 − F = 900a
-100 − F = 900a
Sum of forces on the trailer:
∑F = ma
-100 − 100 = 300a
-200 = 300a
a = -⅔
Plugging into the first equation:
-100 − F = 900 (-⅔)
-100 − F = -600
F = 500 N
, and side AC (x using the special right triangle) is equal to 16. We can therefore say that side BC is equal to 
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