Explanation:
I see an object that is has a Positve acceleration for 4 seconds then a constant acceleration for the final 6 seconds.
ii. Acceleration is the slope of a velocity versus time graph.
Since the first phase is a line, it has a uniform slope.
So we need to chose two points to find the acceleration
Let's use (2,60) and (4,90).
![a = \frac{90 - 60}{4 - 2} = \frac{30}{2} = 15](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7B90%20-%2060%7D%7B4%20-%202%7D%20%20%3D%20%20%5Cfrac%7B30%7D%7B2%7D%20%20%3D%2015)
So the acceleration is 15.
This means, for any velocity versus time graph, the velocity is
![\frac{v _{f} - v _{i}}{t} = a](https://tex.z-dn.net/?f=%20%5Cfrac%7Bv%20_%7Bf%7D%20-%20v%20_%7Bi%7D%7D%7Bt%7D%20%20%3D%20a)
This is velocity final- velocity initial divided by time equals Acceleration
iii. The average velocity is the the sum of the final velocity and initial velocity divided by 2.
So the average velocity is
![\frac{30 + 90}{2} = 60](https://tex.z-dn.net/?f=%20%5Cfrac%7B30%20%2B%2090%7D%7B2%7D%20%20%3D%2060)
Then to find distance
![v _{avg}t = d](https://tex.z-dn.net/?f=v%20_%7Bavg%7Dt%20%3D%20d)
So during the first interval, the motorbike travels
![60 \times (4) = 240](https://tex.z-dn.net/?f=60%20%5Ctimes%20%284%29%20%3D%20240)
In the next part, the average velocity stays at 90 and this occurs for 6 seconds
![90 \times 6 = 540](https://tex.z-dn.net/?f=90%20%5Ctimes%206%20%3D%20540)
Next, we add the velocities
![240 + 540 = 780](https://tex.z-dn.net/?f=240%20%2B%20540%20%3D%20780)
So the distance travleed is 780 meters which is close to 800