Amotorcycle rider moving with an initial velocity of 9.2 m/s uniformly accelerates to a speed of 19.1 m/s in a distance of 32.0 m.
what is the acceleration?

4.38 m/s^2

Explanation:

If,

Initial velocity(u) is 9.2 m/s

Final velocity (v) is 19.1 m/s

Distance moved(s) in that period is 32 m

Therefore we use the formula,

v^2 = u^2 + 2as

; (19.1)^2 = (9.2)^2 + 2(32)a...then move the variables without a to one side..,

; 280.17 = 64a...then divide both sides by 64

; Acceleration = 4.38 m/s^2