Starting from rest a vehicle accelerates at the rate of 2 m/s² towards east for 10 s. It then stops suddenly. It then accelerates again at a rate of 4√2 m/s² for next 10 s towards south and then again comes to rest. The net displacement of the vehicle from the starting point is
- A.100 m
- B.200 m
- C.300 m
- D.400 m
Correct answer
Explanation
The correct answer is C, 300 m. The vehicle goes 100 m east and 200√2 m south, and the straight-line distance between start and finish is √(100² + (200√2)²) = 300 m.
Each leg starts from rest, so the distance covered is s = ½at². Towards the east: s = ½ × 2 × 10² = 100 m. Towards the south: s = ½ × 4√2 × 10² = 200√2 m, which is about 283 m. East and south are at right angles, so the net displacement is the hypotenuse of a right triangle: √(100² + (200√2)²) = √(10,000 + 80,000) = √90,000 = 300 m. Displacement is the shortest distance between the start and end points, not the total path.
A is wrong because 100 m is only the eastward leg.
B is wrong because 200 m ignores the √2 factor in the southward acceleration and adds nothing for the east leg.
D is wrong because 400 m is close to the total path length (about 383 m), not the displacement.
Exam tip: from rest, s = ½at²; perpendicular legs combine by Pythagoras, √(x² + y²).
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