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
<u>Answer</u>
3.44 m/s
<u>Explanation</u>
The motion apply the equations of Newton's law of motion. The ball is acceleration is -9.8 m/s² (acceleration due to gravity. It is negative because the ball is going against gravity, so it is decelerating).
The first equation of Newton's law of motion is;
V = U + at
Where V is the final velocity, U is the initial velocity, a is acceleration and t the time taken.
V = 25 + (-9.8 × 2.2)
= 25 - 21.56
= 3.44 m/s
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