Two motorcycles are traveling due east with different velocities. However, 2.16 seconds later, they have the same velocity. During this 2.16-second interval, motorcycle A has an average acceleration of 1.30 m/s2 due east, while motorcycle B has an average acceleration of 19.4 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 2.16-second interval, and (b) which motorcycle was moving faster?
To solve this problem, we need to use the equations of motion. Let's denote the initial velocities of motorcycle A and motorcycle B as V₀A and V₀B, respectively.
(a) To find the difference in speeds at the beginning of the 2.16-second interval, we can use the formula for calculating velocity:
Velocity (V) = Initial Velocity (V₀) + (Acceleration (a) × Time (t))
For motorcycle A:
V₀A = V₀A + (1.30 m/s² × 2.16 s)
Simplifying, we find:
V₀A = V₀A + 2.808 m/s
Similarly, for motorcycle B:
V₀B = V₀B + (19.4 m/s² × 2.16 s)
Simplifying, we find:
V₀B = V₀B + 41.904 m/s
Now, let's subtract the initial velocities to find the difference in speeds:
Difference in speeds = V₀B - V₀A
Difference in speeds = (V₀B + 41.904 m/s) - (V₀A + 2.808 m/s)
Difference in speeds = V₀B - V₀A + 39.096 m/s
Therefore, the speeds differed by 39.096 m/s at the beginning of the 2.16-second interval.
(b) To determine which motorcycle was moving faster, we compare their initial velocities (V₀A and V₀B).
If V₀A > V₀B, then motorcycle A was moving faster at the beginning of the 2.16-second interval.
If V₀A < V₀B, then motorcycle B was moving faster at the beginning of the 2.16-second interval.
Since we don't have specific values for V₀A and V₀B in the problem statement, we cannot determine which motorcycle was moving faster.
1.30 m/s^2 = (Vt - Vo) / 2.16 s,
Cross multiply:
Vt - Vo = 2.81 m/s,
Delta V = 2.81 m/s for Motorcycle A.
19.4 m/s^2 = (Vt - Vo) / 2.16 s,
Vt - Vo = 41.9 m/s ,
Delta V = 41.9 m/s for Motorcycle B.
41.9 m/s - 2.81 m/s = 39.1 m/s =
Difference in speed at beginning of
2.16 s interval.