Question

50 POINTS!

Question:

Two buses A and B are at positions 60 m and 160 m from the origin at time t = U They start moving in the same direction simultaneously with uniform speed 54 km h-¹ and 36 km h-¹. Determine the time and position at which A overtaken B.


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Answer 1

20 seconds

360 meters

Step-by-step 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:

`\implies \sf 54\; km/h=(54)/(3.6)=15\:m/s`

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:

`\boxed{y=15x+60}`

Bus B

Convert Bus B's given speed of 36 km/h into m/s by dividing by 3.6:

`\implies \sf 36\; km/h=(36)/(3.6)=10\:m/s`

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:

`\boxed{y=10x+160}`

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:

`\implies 15x+60=10x+160`

`\implies 15x+60-10x=10x+160-10x`

`\implies 5x+60=160`

`\implies 5x+60-60=160-60`

`\implies 5x=100`

`\implies (5x)/(5)=(100)/(5)`

`\implies x=20`

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:

`\implies y=10(20)+160`

`\implies y=200+160`

`\implies y=360`

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.