Final answer:
The average acceleration vector of a car whose velocity changes from (15m/s)ît + (8m/s)êj to (-10m/s)ît - (2m/s)êj over 5 seconds is -5m/s²ît - 2m/s²êj.
Step-by-step explanation:
To find the average acceleration vector of a car when its velocity changes from v1 to v2 over a time interval of t, we use the following formula:
a = Δv / Δt
Where:
- Δv is the change in velocity (v2 - v1)
- Δt is the change in time (t seconds)
In this case, the initial velocity v1 is (15m/s)ît + (8m/s)êj and the final velocity v2 is (-10m/s)ît - (2m/s)êj. The change in
velocity Δv is therefore:
Δv = v2 - v1 = ((-10m/s)ît - (2m/s)êj) - ((15m/s)ît + (8m/s)êj)
= (-10 - 15)m/sît - (2 - 8)m/sêj
= -25m/sît - 10m/sêj
Given the time Δt is 5 seconds, the average acceleration a is then calculated as:
a = Δv / Δt = (-25m/sît - 10m/sêj) / 5s = -5m/s²ît - 2m/s²êj
So, the average acceleration vector is -5m/s²ît - 2m/s²êj.