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 t is time taken to cover this distance, applicable kinematic expression is

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