What is the average acceleration (in km/h2) of the bus for the entire 3.5-h period? Point A = (32,1) B = (46, 2.25) C = (38, 3.5)
To find the average acceleration of the bus for the entire 3.5-hour period, we need to calculate the change in velocity over time.
First, let's calculate the change in velocity between points A and B.
Change in velocity (Δv) = v2 - v1
v1 = 1 km/h (from point A)
v2 = 2.25 km/h (from point B)
Δv = 2.25 km/h - 1 km/h = 1.25 km/h
Next, let's calculate the change in velocity between points B and C.
v1 = 2.25 km/h (from point B)
v2 = 3.5 km/h (from point C)
Δv = 3.5 km/h - 2.25 km/h = 1.25 km/h
Now, we have the change in velocity for each leg of the journey. To find the total change in velocity, we add these two changes together.
Total change in velocity (Δv total) = Δv AB + Δv BC
Δv total = 1.25 km/h + 1.25 km/h = 2.5 km/h
Finally, we can calculate the average acceleration using the formula:
Average acceleration = Δv total / time
In this case, the time is 3.5 hours.
Average acceleration = 2.5 km/h / 3.5 hours
= 0.714 km/h²
Therefore, the average acceleration of the bus for the entire 3.5-hour period is 0.714 km/h².