A body moving in a straight line with constant accelaration takes 3 seconds and 5 seconds to cover two successive distances of 1m.Find the accelaration?
It looks straight forward to me. Take the initial speed to be v and acceleration to be a. Take t= 0 at the beginning of the first second in which the object goes 3 feet.
The distance such an object covers is t seconds is [itex]vt+ (1/2)at^2[/itex]. In one second, it will have gone distance v+ (1/2)a. In two seconds it will have gone distance 2v+ 2a. What distance did it go the second second? Set those equal to 3 and 5 and you have two equations to solve for v and a.