A car of mass 800 kg is travelling at a speed of 40 km/h on a straight road. A force of 1000 N is applied on it in the direction of its velocity for 12 s. Which one among the following is the final speed of the car?
- A.55 km/h
- B.60 km/h
- C.94 km/h
- D.104 km/h
Correct answer
Explanation
The correct answer is C, 94 km/h. The force gives the car an extra 15 m/s, which is 54 km/h, and 40 + 54 = 94 km/h. By Newton's second law, acceleration a = F/m = 1000 N ÷ 800 kg = 1.25 m/s². Acting for 12 s, the force raises the speed by a × t = 1.25 × 12 = 15 m/s. The same result follows from impulse: F × t = 1000 × 12 = 12,000 N s, which equals the change in momentum, so the change in speed is 12,000 ÷ 800 = 15 m/s. To convert, multiply by 18/5: 15 × 3.6 = 54 km/h. Because the force acts along the direction of motion, this gain simply adds to the starting 40 km/h. Option A is wrong because 55 km/h would mean a gain of only 15 km/h, which mixes up m/s and km/h. Option B is wrong because 60 km/h does not match any correct step. Option D is wrong because 104 km/h would need a gain of 64 km/h, more than the force provides. Exam tip: change in speed = Ft/m; to turn m/s into km/h multiply by 18/5 (3.6).
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