Given: Change of x is 35.4m, Velocity Final=7.10 m/s, Velocity Initial=0m/s Find: Acceleration Analysis: Vf²=Vi²+2aΔx (Velocity final squared equals Velocity initial squared plus 2 times acceleration times change of x) (7.10 m²/s)²=(0 m/s)²+2a(35.4 m) 50.41 m/s²=(70.8 m)a a=0.712 m/s²
A.No. Assuming no other factors (such as air resistance) the first object will have a velocity of 32 feet/second when the other is dropped. Since they will both have the same acceleration, the first distance between them will increase by 32 feet per second.
