Question

A dog accelerates at 1.50 m/s2 to reach a velocity of 13.5 m/s while covering a distance of 49.3 m. What was his initial velocity?

Answer 1

Let v be the dog's initial velocity. Then

(13.5 m/s)^2 - v ^2 = 2 (1.50 m/s^2) (49.3 m)

==> v ^2 = (13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m)

==> v = √((13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m))

==> v ≈ 5.86 m/s