Respuesta :
Answer:
B. 1400 meters.
Step-by-step explanation:
According to the statement, there are two stages for the motion of the tractor, in which we need to find the travelled distance:
(i) Uniform accelerated motion.
[tex]s_{i} = s_{o,i} +v_{o,i}\cdot t_{i} + \frac{1}{2}\cdot a\cdot t_{i}^{2}[/tex] (1)
[tex]v_{i} = v_{o,i}+a\cdot t_{i}[/tex] (2)
Where:
[tex]s_{o,i}[/tex] - Initial position of the tractor, measured in meters.
[tex]s_{i}[/tex] - Final position of the tractor, measured in meters.
[tex]v_{o,i}[/tex] - Initial velocity of the tractor, measured in meters per second.
[tex]t_{i}[/tex] - Time, measured in seconds.
[tex]a[/tex] - Acceleration, measured in meters per square second.
[tex]v_{i}[/tex] - Final velocity of the tractor, measured in meters per second.
(ii) Uniform motion.
[tex]s_{ii} = s_{i} + v_{i}\cdot t_{ii}[/tex] (3)
Where:
[tex]s_{ii}[/tex] - Final position of the tractor, measured in meters.
[tex]t_{ii}[/tex] - Time, measured in seconds.
The distance covered by the tractor ([tex]\Delta s[/tex]), measured in meters, is:
[tex]\Delta s = s_{ii}[/tex] (4)
If we know that [tex]a = 4\,\frac{m}{s^{2}}[/tex], [tex]t_{i} = 10\,s[/tex], [tex]v_{o,i} = 0\,\frac{m}{s}[/tex], [tex]s_{o,i} = 0\,m[/tex] and [tex]t_{ii} = 30\,s[/tex], then distance covered by the tractor is:
By (1) and (2):
[tex]s_{i} = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (10\,s)+ \frac{1}{2}\cdot \left(4\,\frac{m}{s^{2}} \right) \cdot (10\,s)^{2}[/tex]
[tex]s_{i} = 200\,m[/tex]
[tex]v_{i} = \left(0\,\frac{m}{s} \right)+\left(4\,\frac{m}{s^{2}} \right)\cdot (10\,s)[/tex]
[tex]v_{i} = 40\,\frac{m}{s}[/tex]
By (3):
[tex]s_{ii} = 200\,m + \left(40\,\frac{m}{s} \right)\cdot (30\,s)[/tex]
[tex]s_{ii} = 1400\,m[/tex]
By (4):
[tex]\Delta s = 1400\,m[/tex]
The correct answer is B.