Physics · Motion In A Straight Line · NEET
No, and this is the most common mistake. Distance in n seconds is the total path from the start (t = 0) up to time t = n, given by S(n) = un + (1/2)an^2. Distance in the nth second is only the small distance covered during that single second, from t = n-1 to t = n. You get it by subtracting: S(nth) = S(n) - S(n-1). For example, distance in 3 seconds is much larger than distance in the 3rd second alone.
Because the second is understood as a factor of 1 second, so the formula secretly means [distance per 1 second] x (1 s). The term u + a(2n-1)/2 numerically equals the average velocity during that one second (in m/s), and multiplying by the 1 second time interval gives a distance in metres. So the answer is a distance (metre), not a speed. Never write the unit as m/s for S(nth).
Start with S = un + (1/2)an^2. Distance in the nth second = S(n) - S(n-1) = [un + (1/2)an^2] - [u(n-1) + (1/2)a(n-1)^2]. Expand: un - u(n-1) = u. And (1/2)a[n^2 - (n-1)^2] = (1/2)a[n^2 - n^2 + 2n - 1] = (1/2)a(2n - 1). Add them: S(nth) = u + (a/2)(2n - 1). That is the standard result.
Put u = 0. Then S(nth) = (a/2)(2n - 1), which is proportional to (2n - 1) = 1, 3, 5, 7... This gives the famous 1:3:5:7 ratio of distances in successive seconds. For free fall, a = g = 10 m/s^2 (NEET usually uses 10), so 1st second = 5 m, 2nd second = 15 m, 3rd second = 25 m, and so on.
Yes, if the acceleration is opposite to the motion (retardation) and n is large, the (2n-1) term with negative a can make S(nth) come out negative. A negative value means during that particular second the body moved backward (opposite to the original positive direction), which happens after it has stopped and reversed. Always keep the sign convention consistent for a.
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.
S(nth) = u + a(2n - 1)/2, where u is the initial velocity, a is the uniform acceleration, and n is the number of the second (like 1st, 2nd, 3rd...). It is valid only for uniformly accelerated (constant a) motion.
Metre (m). Even though the formula looks like a velocity expression, the time interval of 1 second is implied, so the result is a distance. Do not write it as m/s.
Yes. Just set u = 0, giving S(nth) = (a/2)(2n - 1). This produces the 1:3:5:7 ratio for successive seconds, a very common NEET result for free fall and objects released from rest.
Distance in n seconds, S(n) = un + (1/2)an^2, is the full path up to time n. Distance in the nth second, S(nth) = u + a(2n-1)/2, is only the slice covered during that one second, equal to S(n) - S(n-1).
Yes. Replace a with g (take a = g for downward fall, usually 10 m/s^2 in NEET). For a body dropped from rest, u = 0, so S(nth) = (g/2)(2n - 1).
