A motorcycle 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