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#
where#v# is final velocity,#u# is the initial velocity,#a# is acceleration and#s# 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
