Distance covered in last second
Three particles move in a straight line with initial velocities v 1 , v 2 and v 3 ( v 1 < v 2 < v 3 ) each with constant retardation ' a ' such that motion continues for more than one second before velocity of each particle becomes zero. If s 1 , s 2 and s 3 respectively be the distances travelled in the last one second before velocity becomes zero, then
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2 solutions
Let us consider the first particle. Time of travel before coming to rest is t = a v 1 .
Distance traveled during the last second is s 1 = v 1 − 2 a ( 2 t − 1 ) = 2 a . Hence this is independent of the initial velocity of the particle, and is the same for all the particles, since acceleration is the same for all of them.
If the acceleration of the particles are constant in time and uniform across the group then they must ALL experience the same change in velocity per unit time [1].
Regardless of the initial velocities of the particles they ALL end with the same velocity, 0 [T] [L] . That means that in addition to ALL having the same acceleration through the duration of motion, 1 second prior to having no velocity they ALL had the same velocity as well due to [1]. Thus they ALL traveled the same distance in their final second before zero velocity was reached. s 1 = s 2 = s 3