Answer:
The final displacement of the car is 140 meters.
Explanation:
The final displacement of the car (
), in meters, is the sum of the change in displacement associated with each part of the journey, which is derived from the following kinematic formulas:
(1)
(2)
(3)
(4)
Where:
- Traveled distance of the first part, in meters.
- Traveled distance of the second part, in meters.
- Acceleration in the first part, in meters per square second.
- Acceleration in the second part, in meters per square second.
- Initial speed of the car in the second part, in meters per second.
- Time taken in the first part, in seconds.
- Time taken in the second part, in seconds.
If we know that
,
,
and
, then the distance traveled by the car is:
By (2):
![s_{1} = \frac{1}{2}\cdot \left(0.8\,\frac{m}{s^{2}} \right)\cdot(10\,s)^{2}](https://tex.z-dn.net/?f=s_%7B1%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%280.8%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29%5Ccdot%2810%5C%2Cs%29%5E%7B2%7D)
![s_{1} = 40\,m](https://tex.z-dn.net/?f=s_%7B1%7D%20%3D%2040%5C%2Cm)
By (3):
![v_{o,2} = \left(0.8\,\frac{m}{s^{2}} \right)\cdot (10\,s)](https://tex.z-dn.net/?f=v_%7Bo%2C2%7D%20%3D%20%5Cleft%280.8%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29%5Ccdot%20%2810%5C%2Cs%29)
![v_{o,2} = 8\,\frac{m}{s}](https://tex.z-dn.net/?f=v_%7Bo%2C2%7D%20%3D%208%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
By (4):
![s_{2} = \left(8\,\frac{m}{s} \right)\cdot (10\,s) + \frac{1}{2}\cdot \left(0.4\,\frac{m}{s^{2}} \right)\cdot (10\,s)^{2}](https://tex.z-dn.net/?f=s_%7B2%7D%20%3D%20%5Cleft%288%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5Ccdot%20%2810%5C%2Cs%29%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%280.4%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29%5Ccdot%20%2810%5C%2Cs%29%5E%7B2%7D)
![s_{2} = 100\,m](https://tex.z-dn.net/?f=s_%7B2%7D%20%3D%20100%5C%2Cm)
By (1):
![s = 140\,m](https://tex.z-dn.net/?f=s%20%3D%20140%5C%2Cm)
The final displacement of the car is 140 meters.