Motion in a Straight Line (NEET)

Question 1

A body moving along a straight line with uniform acceleration covers 10 m in the 3rd second and 18 m in the 5th second of its motion. The distance it covers in the 6th and 7th seconds together is

Accepted Answer

C. Key insight: sₙ = u + a/2(2n − 1) is linear in n, so two values fix u and a. u + 2.5a = 10 and u + 4.5a = 18 give a = 4 m/s² and u = 0. Distance in 6th second = 0 + 2(11) = 22 m and in 7th second = 2(13) = 26 m; total 48 m. Check: s(7) − s(5) = 2(49) − 2(25) = 48 m. Trap: 26 m is the 7th second alone; 44 m and 52 m come from off-by-one errors in 2n − 1. Exam pattern: NEET – distance travelled in t

Question 2

A ball dropped from rest from the top of a tower covers 9/25 of the tower's height in the last second of its fall. Take g = 10 m/s². The height of the tower is

Accepted Answer

D. Key insight: with total fall time n seconds, the fraction covered in the last second is ((2n−1))/(n²), since sₙ = g/2(2n−1) and H = g/2n². So (2n − 1)/(n²) = 9/25, that is 9n² − 50n + 25 = 0, giving n = 5 or n = 5/9. The root 5/9 s is rejected because the fall must last more than the 1 s mentioned. Hence H = 5 × 25 = 125 m. Trap: keeping the spurious root, or using H = 1/2gn style slips leading to

Question 3

A car brakes with a uniform deceleration and comes to rest. It covers 20 m in the first second of braking and 4 m in the last second before stopping. The total distance covered while braking is

Accepted Answer

A. Key insight: the last second before stopping is the time-reverse of the first second of motion from rest, so it fixes the deceleration directly. Last second: 1/2a(1)² = 4, so a = 8 m/s². First second: u − a/2 = 20, so u = 24 m/s. Total distance = (u²)/(2a) = 576/16 = 36 m (braking time 3 s). Trap: adding 20 + 4 or taking u = 20 m/s (400/16 = 25 m is not even listed; 24 m and 40 m come from similar

Motion in a Straight Line: questions with solutions

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