Acar accelerates uniformly from rest and reaches a speed of 18 m/s in 13.8 s. the diameter of a tire is 60.8 cm. find the number of revolutions the tire makes during this motion, assuming no slipping.

The area under a velocity time graph is distance travelled (as integrating velocity gives distance), so we can imagine the v/t as a line, giving a triangle with 2 sides of 13.8 and 18. The area of a triangle is bh/2, so 9*13.8 = 124.2m which is how far the car travelled in that time.
60.8cm is 0.608m, and now we can divide to work out the number of revolutions:
124.2/0.608 = 204.28 = 204 revolutions to 3sf (number of sf in the question assuming 18 is in fact 18.0, so the number you should give your answer to)

    5.5 n======    assuming that      we have      m = 0.145 kg    a = 38 m/s²therefore    5.51 n, but two significant figures.final answer: 5.5 n

This is the answer: body 1 + engine 2