Question

A fish swimming in a horizontal plane has velocity i = (4.00 + 1.00 ) m/s at a point in the ocean where the position relative to a certain rock is i = (12.0 ? 1.80 ) m. After the fish swims with constant acceleration for 19.0 s, its velocity is = (17.0 ? 1.00 ) m/s.
(a) What are the components of the acceleration of the fish?
ax =

m/s2
ay = -0.10


Review the definition of average acceleration and remember that each component is treated separately. m/s2

(b) What is the direction of its acceleration with respect to unit vector ?

Answer 1

Answer:

a. The horizontal component of acceleration a₁ = 0.68 m/s²

The vertical component of acceleration a₂ = -0.11 m/s²

b. -9.19° = 350.81° from the the positive x-axis

Explanation:

The initial velocity v₁ of the fish is v₁ = 4.00i + 1.00j m/s. Its final velocity after accelerating for t = 19.0 s is v₂ =  17.0i - 1.00j m/s

a. The acceleration a = (v₂ - v₁)/t = [17.0i - 1.00j - (4.00i + 1.00j)]/19 = [(17.0 -4.0)i - (-1.0 -1.0)j]/19 = (13.0i - 2.0j)/19 = 0.68i - 0.11j m/s²

The horizontal component of acceleration a₁ = 0.68 m/s²

The vertical component of acceleration a₂ = -0.11 m/s²

b. The direction of the acceleration relative to the unit vector i,

tanθ = a₂/a₁ = -0.11/0.68 = -0.1618

θ = tan⁻¹(-0.1618) = -9.19° ⇒ 360 + (-9.19) = 350.81° from the the positive x-axis