An object with an initial speed of 50.0 ft/s is accelerating at 2.00ft/s^2. Find (a) the speed after the body has covered a distance of 500ft and (b) the time taken to cover this distance?
1 Answer
Jul 4, 2018
(a) The kinematic expression which can be used is
v^2-u^2=2as
wherev is final velocity,u is the initial velocity,a is acceleration ands is displacement
Inserting given values we get
v^2-(50.0)^2=2(2.00)500
=>v^2=2(2.00)500+(50.0)^2
=>v=sqrt(2(2.00)500+(50.0)^2)
=>v=67.1\ ft\ s^-1 , rounded to one decimal place
(b) If
v=u+at
Inserting given and calculated values we get
67.08=50.0+2.00xxt
=>t=(67.08-50.0)/2.00
=>t=8.5\ s , rounded to one decimal place