Answer:
The final displacement of the car is 140 meters.
Explanation:
The final displacement of the car ([tex]s[/tex]), 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:
[tex]s = s_{1} + s_{2}[/tex] (1)
[tex]s_{1} = \frac{1}{2}\cdot a_{1}\cdot t_{1}^{2}[/tex] (2)
[tex]v_{o,2} = a_{1}\cdot t_{1}[/tex] (3)
[tex]s_{2} = v_{o,2} + \frac{1}{2}\cdot a_{2}\cdot t_{2}^{2}[/tex] (4)
Where:
[tex]s_{1}[/tex] - Traveled distance of the first part, in meters.
[tex]s_{2}[/tex] - Traveled distance of the second part, in meters.
[tex]a_{1}[/tex] - Acceleration in the first part, in meters per square second.
[tex]a_{2}[/tex] - Acceleration in the second part, in meters per square second.
[tex]v_{o,2}[/tex] - Initial speed of the car in the second part, in meters per second.
[tex]t_{1}[/tex] - Time taken in the first part, in seconds.
[tex]t_{2}[/tex] - Time taken in the second part, in seconds.
If we know that [tex]a_{1} = 0.8\,\frac{m}{s^{2}}[/tex], [tex]t_{1} = 10\,s[/tex], [tex]a_{2} = 0.4\,\frac{m}{s^{2}}[/tex] and [tex]t_{2} = 10\,s[/tex], then the distance traveled by the car is:
By (2):
[tex]s_{1} = \frac{1}{2}\cdot \left(0.8\,\frac{m}{s^{2}} \right)\cdot(10\,s)^{2}[/tex]
[tex]s_{1} = 40\,m[/tex]
By (3):
[tex]v_{o,2} = \left(0.8\,\frac{m}{s^{2}} \right)\cdot (10\,s)[/tex]
[tex]v_{o,2} = 8\,\frac{m}{s}[/tex]
By (4):
[tex]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}[/tex]
[tex]s_{2} = 100\,m[/tex]
By (1):
[tex]s = 140\,m[/tex]
The final displacement of the car is 140 meters.
