If an object moves at a non-zero constant acceleration for a certain interval of time, then the distance it covers in that time
- (a)depends on its initial velocity.
- (b)is independent of its initial velocity.
- (c)increases linearly with time.
- (d)depends on its initial displacement.
Correct — A, it depends on its initial velocity. For constant acceleration the distance is s = ut + ½at², and the first term ut shows the distance depends on the initial velocity u. Two bodies with the same acceleration but different starting speeds cover different distances in the same interval, so the distance is not independent of u.
- (b)is independent of its initial velocity. — The term ut in s = ut + ½at² makes the distance depend directly on the initial velocity u, so it is not independent of it.
- (c)increases linearly with time. — With non-zero acceleration the distance has a t² term, so it grows quadratically (non-linearly) with time, not linearly.
- (d)depends on its initial displacement. — The distance covered in the interval is measured from the starting point; where that point lies (the initial displacement) does not affect how far the body travels.
The kinematic equation s = ut + ½at² gives the displacement of a uniformly accelerated body during a time t. It depends on both the initial velocity u and the time, but not on the absolute starting position. Because of the ½at² term the distance grows quadratically, not linearly, with time.
Read the answer straight off the equation of motion: the ut term ties the distance to the initial velocity, and the ½at² term makes it non-linear in time. The distractor about initial displacement confuses 'where you start' with 'how far you go'.
- Equation of motion: s = ut + ½at².
- Other forms: v = u + at and v² = u² + 2as.
- With a ≠ 0 the distance varies as t² (quadratic, not linear).
- The ut term makes the distance depend on the initial velocity.
Both the initial velocity and the acceleration set the distance; it rises as t², and the starting position is irrelevant.
- Assuming distance grows linearly with time when acceleration is non-zero — it is quadratic.
- Confusing initial displacement (starting position) with initial velocity (starting speed).
Asked as 'the distance under constant acceleration depends on / is independent of ...' — read it directly from s = ut + ½at².
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 concept — the equation of motion s = ut + ½at². UPSC tests its shape (a free-fall body gives s = ½gt², a parabola in time); this NDA item tests that the distance depends on the initial velocity term ut.
A car has an initial velocity of 12 m/s and is brought to rest over a distance of 45 m. The acceleration of the car is
- (a) +1.6 m/s²
- (b) +3.2 m/s²
- (c) −1.6 m/s²
- (d) −0.8 m/s²
Answer(c) −1.6 m/s²
Same toolkit — the equations of uniform acceleration. That NDA item applies v² = u² + 2as with the initial velocity to find the deceleration; this one tests that the distance itself depends on that initial velocity.
- practice — not a real PYQ
A body starts from rest and moves with uniform acceleration a. The distance covered in time t is
- (a)at
- (b)½at²
- (c)at²
- (d)2at
Answer(b) ½at² — with u = 0, s = ut + ½at² reduces to ½at².
- practice — not a real PYQ
For a uniformly accelerated body, the distance–time graph is
- (a)a straight line
- (b)a parabola
- (c)a horizontal line
- (d)a circle
Answer(b) a parabola — the ½at² term makes distance vary as the square of time.