course ph121

Determine the acceleration of an object which, after accelerating through a distance of 53 cm if 3.4 sec, is moving at 8.17647 cm/s.

2(`ds/`dt)-vF=v0

2(53/3.4)-8.17647=23cm/s

(vF^2 - v0^2)/(2`ds)=A

(8.17647^2/23^2)/(2*53)=-4.36cm/s/s

acceleration of the object is (-4.36cm/s/s)