Question

I need some help. :) A point moves from O in a straight line X’OX with an anitial velocity of 7 m/s. in the positive direction, the acceleration being constant. After 20 seconds, the point is again at O. a) Find the acceleration and the velocity when the point again passes O. b) Find the position of the point when the velocity is zero. c) Find when the displacement is 21 meters on the positive side of O, and when it is 12 meters on the negative side of O.

Answer 1

let acceleration is a, now in eqn \[s=ut+\frac{ 1 }{ 2} a t ^{2}\] put s=0 and t=20 sec you will get acceleration a. now in eqn \[v ^{2} = u ^{2} +2a s\] put s=0 u will get v=u= 7m/s but direction is opposite.
b) again using equation \[v ^{2} = u ^{2} +2a s\] put v=0 and u=7 m/s and value of a u calculate in part a) u will get s.
c) in eqn \[s=ut+\frac{ 1 }{ 2} a t ^{2}\] as u know a and u first put s=21 m then find corresponding t. now put s=-12 the calculate corresponding t. remember in this question a will be negative so put the value of a with negative sign in equns.

Answer 2

Thanks Gyanu. :)

Answer 3

you are welcome Nadett.