A hockey puck, with an initial velocity of 65 km/h [W], ricochets off the boards. After 0.76 s in contact with the boards, its final velocity is 47 km/h [E]. Determine the acceleration of the puck.

 a = 40.937 m / s²

Explanation:

For this exercise let's use the relationship between momentum and momentum variation

          I  = Δp

          F t = m v_f - mv₀

          F = m (v_f -v₀) / t

let's reduce the magnitudes to the SI system

          v_f = 47 km / h (1000m / 1 km) (1h / 3600 s) = 13.056 m / s

          v₀ = - 65 km / h = -18.056 m / s

the negative sign is bearing the speed is west

let's calculate

          F = m (13.056 + 18.056) / 0.76

          F = m 40.937

now we can use Newton's second law

          F = m a

          m 40.937 = m a

          a = 40.937 m / s²

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2