Starting from rest a vehicle accelerates at the rate of 2 m/s^2 towards east for 10 s. It then stops suddenly. It then accelerates again at a rate of 4 sqrt(2) m/s^2 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 — C, 300 m. East leg: starting from rest, s1 = (1/2) x a x t^2 = (1/2) x 2 x 10^2 = 100 m east. The vehicle stops, then the south leg also starts from rest: s2 = (1/2) x (4 sqrt 2) x 10^2 = 200 sqrt 2 m, about 282.8 m south. The two displacements are perpendicular, so the net straight-line displacement is the hypotenuse: sqrt(100^2 + (200 sqrt 2)^2) = sqrt(10000 + 80000) = sqrt(90000) = 300 m.
- (a)100 m — Only the eastward leg — it ignores the southward displacement altogether.
- (b)200 m — Mis-handles the south leg (e.g. taking it as 200 m) or subtracts the legs, instead of combining two perpendicular displacements.
- (d)400 m — This is roughly the arithmetic sum 100 + 283 rounded up — adding the legs directly instead of combining them as perpendicular vectors.
Each leg is motion from rest under constant acceleration, so its length is s = (1/2) a t^2. The eastward and southward displacements are at right angles, and displacement is a vector, so the net displacement is found by the Pythagorean combination R = sqrt(x^2 + y^2), not by adding the two distances.
The trap is to add the two legs as plain numbers (100 + about 283, giving roughly 383 or 400) instead of combining them at right angles. The figures are chosen so the perpendicular resultant is a clean 300 m — a right triangle with legs 100 and 200 sqrt(2).
- From rest, displacement s = (1/2) x a x t^2.
- East leg: (1/2) x 2 x 10^2 = 100 m; south leg: (1/2) x 4 sqrt(2) x 10^2 = 200 sqrt(2), about 282.8 m.
- Perpendicular displacements combine as vectors: R = sqrt(x^2 + y^2).
- sqrt(100^2 + (200 sqrt 2)^2) = sqrt(10000 + 80000) = sqrt(90000) = 300 m.
- Adding perpendicular displacements arithmetically instead of as vectors.
- Forgetting that the vehicle restarts from rest for the second leg.
Asked as a two-leg motion problem where the perpendicular displacements must be combined as vectors, not added.
The variation of displacement (d) with time (t) in the case of a particle falling freely under gravity from rest is correctly represented by which of the following graphs?
- (a) graph (a)
- (b) graph (b)
- (c) graph (c)
- (d) graph (d)
Answer(a) graph (a)
Same kinematics idea of displacement under constant acceleration from rest — s = (1/2) a t^2 (a parabola in d-t) — which this NDA item applies to each leg before combining the two perpendicular displacements as vectors.
- practice — not a real PYQ
A body moves 3 m towards east and then 4 m towards north. Its displacement from the starting point is
- (a)7 m
- (b)5 m
- (c)1 m
- (d)12 m
Answer(b) 5 m — sqrt(3^2 + 4^2) = 5, the perpendicular resultant.
- practice — not a real PYQ
Starting from rest and accelerating at 2 m/s^2 for 5 s, the distance covered by a body is
- (a)10 m
- (b)25 m
- (c)50 m
- (d)5 m
Answer(b) 25 m — s = (1/2) x 2 x 5^2 = 25 m.