For the velocity-time graph given, the average accelerations during the time intervals 0.0 s to 5.0 s, 5.0 s to 15.0 s, and 0.0 s to 20.0 s are 0, 1.3 m/s², and 0.65 m/s², respectively.
The average acceleration can be calculated with the following equation:
[tex] \overline{a} = \frac{\Delta v}{\Delta t} = \frac{v_{2} - v_{1}}{t_{2} - t_{1}} [/tex] (1)
Where:
v: is the velocity
t: is the time
Now, let's find the average acceleration for each time interval
t₁ = 0.0 s
t₂ = 5.0 s
v₁ = -6.5 m/s (approximately)
v₂ = -6.5 m/s (approximately)
Hence, the average acceleration (eq 1) is:
[tex] \overline{a} = \frac{v_{2} - v_{1}}{t_{2} - t_{1}} = \frac{-6.5 m/s - (-6.5 m/s)}{5.0 s - 0.0 s} = 0 [/tex]
The average acceleration in this interval is zero because the velocity is constant.
- Time interval 5.0 to 15.0
t₁ = 5.0 s
t₂ = 15.0 s
v₁ = -6.5 m/s (approximately)
v₂ = 6.5 m/s (approximately)
Then, the average acceleration (eq 1) is:
[tex] \overline{a} = \frac{6.5 m/s - (-6.5 m/s)}{15.0 s - 5.0 s} = 1.3 m/s^{2} [/tex]
We can see an increasing average acceleration of 1.3 m/s² in this interval.
- Time interval 0.0 to 20.0
t₁ = 0.0 s
t₂ = 20.0 s
v₁ = -6.5 m/s (approximately)
v₂ = 6.5 m/s (approximately)
Therefore, the average acceleration (eq 1) is:
[tex] \overline{a} = \frac{6.5 m/s - (-6.5 m/s)}{20.0 s - 0.0 s} = 0.65 m/s^{2} [/tex]
In this interval, we have an increasing average acceleration of 0.65 m/s², lower than the second interval.
Therefore, the average accelerations during the time intervals 0.0 s to 5.0 s, 5.0 s to 15.0 s, and 0.0 s to 20.0 s are 0, 1.3 m/s², and 0.65 m/s², respectively.
Learn more about average acceleration here https://brainly.com/question/104491?referrer=searchResults
I hope it helps you!
