Physics · Motion In A Straight Line · NEET
1:2:3:4 would mean the distance grows by the same amount each second — that is uniform velocity, not acceleration. With acceleration the body speeds up, so each second it covers MORE extra distance than the last. The distance in the nth second is S_n = u + (a/2)(2n - 1). Starting from rest (u = 0), S_n is proportional to (2n - 1), which gives 1, 3, 5, 7 — the odd numbers. The gap between them is a constant 2, which is exactly why acceleration is constant.
They answer two different questions. 1:3:5:7 is the ratio of distances covered DURING each separate second (1st second alone, 2nd second alone, 3rd second alone). 1:4:9:16 is the ratio of TOTAL distance from the start up to the end of 1s, 2s, 3s, 4s (because total distance from rest is S = (1/2)at^2, so S is proportional to t^2 = 1, 4, 9, 16). Read the question carefully: 'in the nth second' means 1:3:5; 'in n seconds' or 'total' means 1:4:9.
No. It works for ANY uniformly accelerated motion that starts from rest — a car accelerating from a stop, a block sliding from rest on an incline, or a freely falling body. Free fall is just the most common NEET example because a = g. The value of a (or g) cancels out when you take the ratio, so the ratio is always 1:3:5:7 regardless of the acceleration, as long as the body starts from rest.
Then the ratio is no longer 1:3:5:7. You must use the full formula S_n = u + (a/2)(2n - 1) for each second and compute the actual distances, then take the ratio. For example, with u = 5 m/s and a = 2 m/s^2: S_1 = 5 + 1(1) = 6, S_2 = 5 + 1(3) = 8, S_3 = 5 + 1(5) = 10. The ratio 6:8:10 = 3:4:5, not 1:3:5. The neat odd-number ratio is a special result only for u = 0.
The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second of its motion is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It states that a body starting from rest under uniform acceleration covers distances in the ratio 1 : 3 : 5 : 7 ... in the 1st, 2nd, 3rd, 4th ... second. These are the odd numbers, coming from S_n proportional to (2n - 1).
S_n = u + (a/2)(2n - 1), where u is initial velocity, a is acceleration and n is the second number. Note S_n has units of metres (distance in one second), not m/s, even though it looks like a velocity.
1 : 3 : 5. Starting from rest, S_n is proportional to (2n - 1), giving 1, 3, 5 for n = 1, 2, 3.
No. Total distances after 1, 2, 3 seconds are in the ratio 1 : 4 : 9 (proportional to t^2). The 1 : 3 : 5 ratio is only for distance covered during each separate second.
No. The acceleration cancels out when you take the ratio. Whether it is free fall (a = g) or a car (a = 2 m/s^2), a body from rest always gives 1 : 3 : 5 : 7.
