A car which is traveling at a velocity of 9.6 m/s undergoes an acceleration of 4.2 m/s2 over a distance of 450 m. How fast is it going after that acceleration?

Answer 1

d = Vi t + (1/2) a t^2
450 = 9.6 t + 2.1 t^2
solve quadratic for t
then
v = Vi + a t
v = 9.6 + 4.2 t