Question

A man crosses a 90m long straight track with a uniform acceleration in 6 s . If his initial velocity is 3m/s , then he leaves the track with velocity?

Answer 1

s= 90m

t= 6 s

u= 3

using second equation of motion that is S=ut +at²/2

90= 18+ 18a

  a=4m/s²

using first equation of motion that is v=u+at

v= 3 +24= 27m/s