Final answer:
The components of the fish's acceleration are 0.8i - 0.3j m/s². The direction of the acceleration with respect to the unit vector ı is 0.8ı. At t = 25 s, the fish is at a position of 520i - 72.75j m and is moving in the direction of -72.75j.
Step-by-step explanation:
A)
The components of the acceleration of the fish can be determined by subtracting the initial velocity from the final velocity and then dividing by the time taken. The initial velocity vi = 4i + j m/s, and the final velocity vf = 20i - 5j m/s. The time taken is 20 s.
Thus,
Acceleration = (vf - vi)/t
= (20i - 5j - 4i - j)/20
= (16i - 6j)/20
= 0.8i - 0.3j m/s²
B)
The direction of the acceleration with respect to the unit vector ı can be determined by taking its dot product with ı, which gives the scalar component of the acceleration in the direction of ı. The result is then multiplied by ı to obtain the direction of the acceleration vector.
Thus,
Direction of acceleration with respect to ı = (acceleration • ı) * ı
= (0.8i - 0.3j • ı) * ı
= (0.8 * 1) * ı
= 0.8ı
C)
If the fish maintains constant acceleration, its final position and direction of movement can be determined using the equations of motion.
Given that the initial position ri = 10i - 4j m and the time taken t = 25 s, we can use the equation:
Position = Initial position + Initial velocity * time + 0.5 * acceleration * time²
= (10i - 4j) + (4i + j) * 25 + 0.5 * (0.8i - 0.3j) * 25²
= (10i - 4j) + (4i + j) * 25 + 0.4i * 625 - 0.15j * 625
= (10 + 100 + 400 + 10) i + (-4 + 25 - 93.75) j
= 520i - 72.75j m
Therefore, at t = 25 s, the fish is at a position of 520i - 72.75j m and is moving in the direction of -72.75j.