Answer:
20 seconds
360 meters
Explanation:
On a distance-time graph:
- x-axis = time (in seconds)
- y-axis = distance (in meters)
Therefore:
- The starting position (when t = 0) is the y-intercept.
- The speed of the object (in m/s) is the gradient of the line.
Bus A
Convert Bus A's given speed of 54 km/h into m/s by dividing by 3.6:
[tex]\implies \sf 54\; km/h=\dfrac{54}{3.6}=15\:m/s[/tex]
Therefore:
- At t = 0, Bus A is at 60 m from the origin.
- It moves with a uniform speed of 15 m/s.
So the y-intercept is (0, 60) and the gradient is 15.
Substitute this information into the slope-intercept form of a linear equation to create an equation representing the motion of Bus A:
[tex]\boxed{y=15x+60}[/tex]
Bus B
Convert Bus B's given speed of 36 km/h into m/s by dividing by 3.6:
[tex]\implies \sf 36\; km/h=\dfrac{36}{3.6}=10\:m/s[/tex]
Therefore:
- At t = 0, Bus B is at 160 m from the origin.
- It moves with a uniform speed of 10 m/s.
So the y-intercept is (0, 160) and the gradient is 10.
Substitute this information into the slope-intercept form of a linear equation to create an equation representing the motion of Bus B:
[tex]\boxed{y=10x+160}[/tex]
Solution
To find the time when Bus A is at the same position as Bus B, substitute Bus A's equation into Bus B's equation and solve for x:
[tex]\implies 15x+60=10x+160[/tex]
[tex]\implies 15x+60-10x=10x+160-10x[/tex]
[tex]\implies 5x+60=160[/tex]
[tex]\implies 5x+60-60=160-60[/tex]
[tex]\implies 5x=100[/tex]
[tex]\implies \dfrac{5x}{5}=\dfrac{100}{5}[/tex]
[tex]\implies x=20[/tex]
Therefore, the two buses are at the same position 20 seconds after they start moving.
To find their position at this point, substitute x = 20 into one of the equations and solve for y:
[tex]\implies y=10(20)+160[/tex]
[tex]\implies y=200+160[/tex]
[tex]\implies y=360[/tex]
Therefore, the two buses are at the same position at 360 m from the origin.
Conclusion
Bus A overtakes Bus B 20 seconds after they start moving and at 360 m from the origin.
