an ice skater starts out traveling with a velocity of (-3,-7) m/s. he performs a 3 second maneuver and ends with a velocity of (0, 5) m/s. (a) what is his average acceleration over this period? (b) a different ice skater starts with the same initial velocity, accelerates at (1.5, 3.5) m/s2 for 2 seconds, and then at (0, 5 m.s2 for 1 seconds. what is his final velocity?
(a)
(b)
(a) The components of the average acceleration and the changes in the velocity components,(3, 12) divided by the time interval. You do the numbers
(b) Add the two different velocity vector changes during the two time intervals PLUS the initial velocity vector.
You have only specified one component of the acceleration during the last 1 second. Perhaps you did not copy the question correctly.
(a) To find the average acceleration, we need to calculate the change in velocity and divide it by the time taken. The change in velocity is simply the final velocity minus the initial velocity.
Given:
Initial velocity (u) = (-3, -7) m/s
Final velocity (v) = (0, 5) m/s
Time taken (t) = 3 seconds
Change in velocity (Δv) = v - u = (0, 5) - (-3, -7) = (3, 12) m/s
Now, we can calculate the average acceleration using the formula:
Average acceleration (a_avg) = Δv / t
Substituting the values, we get:
a_avg = (3, 12) / 3 = (1, 4) m/s²
Therefore, the average acceleration over this period is (1, 4) m/s².
(b) For the second ice skater, we need to calculate the final velocity after the different accelerations. Let's break down the problem step by step.
Given:
Initial velocity (u) = (-3, -7) m/s
Accelerations:
- First acceleration (a1) = (1.5, 3.5) m/s² for 2 seconds
- Second acceleration (a2) = (0, 5) m/s² for 1 second
To find the final velocity, we need to consider each acceleration separately and apply them sequentially.
1. First, calculate the velocity after the first acceleration (a1) for 2 seconds using the formula:
v1 = u + a1 * t1
Substituting the values, we get:
v1 = (-3, -7) + (1.5, 3.5) * 2 = (-3, -7) + (3, 7) = (0, 0) m/s
2. Next, calculate the velocity after the second acceleration (a2) for 1 second using the same formula:
v2 = v1 + a2 * t2
Substituting the values, we get:
v2 = (0, 0) + (0, 5) * 1 = (0, 0) + (0, 5) = (0, 5) m/s
Therefore, the final velocity of the second ice skater is (0, 5) m/s.