A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three seconds of its motion. The stone remains in the air for – (a) 6s (b) 5s (c) 7s (d) 4s

By BYJU'S Exam Prep

Updated on: September 25th, 2023

A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three seconds of its motion. The stone remains in the air for 5s. The correct option is B 5s. Given that the stone falls from rest, so u = 0:

The final second’s distance is now equal to the motion’s first three seconds’ worth of travel. Consequently, using the second law of motion, we obtain

d3 = 0 + ½ gt2

= ½ x 10 x 9

= 45

Use Sn = u + ½ a (2n – 1), to find distance covered at any nth second

St = u + (2t – 1) x 10/2

According to the question,

d3 = St

5 (2t – 1) = 45

t = 5s

Table of content

1. A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three seconds of its motion. The stone remains in the air for – (a) 6s (b) 5s (c) 7s (d) 4s

Relationship between speed, time, and distance

Distance is the sum of an object’s movements, regardless of direction. The distance can be defined as the amount of space an object has covered, regardless of its starting or ending position.

Speed = Distance/Time – This reveals how quickly or slowly an object is moving. It gives the amount of time it took to travel a certain distance divided by the distance traveled.

Speed is inversely correlated with time and directly correlated with distance. Hence,

Distance = Speed X Time, and

Time = Distance / Speed, as the speed increases the time taken will decrease and vice versa.

Summary: