Answer:
Part A
The two equations are;
d = 3 m/s × t for Boat A
d = 4.8 m/s × (t - 6 s) for Boat B
Part B
The solution to the system of equation are
t = 16 s and d = 48 m
Part C: The possible values of 't' for Boat A are 0 ≤ t ≤ ∞
The possible values of 't' for Boat B are 6 ≤ t ≤ ∞
Boat B timing starts when Boat B crosses the start line when t = 6 seconds
The solution of the system of equation indicates that Boat B and Boat A arrive at the same point, 'd' 48 meters from the start, 16 seconds from the time Boat A crosses the stat line
Step-by-step explanation:
The given parameters are;
The speed of Boat A = 3 m/s
The speed of Boat B = 4.8 m/s
The time Boat B crosses the start line = 6 seconds after Boat A crosses the start line
Part A
The two equations are;
For Boat A
d = 3 m/s × t...(1)
For Boat B
d = 4.8 m/s × (t - 6 s)...(2)
Where;
t = The time in seconds since Boat A crossed the start line
d = The distance traveled by the boat in meters
Part B
The system of equations from Part A is solved as follows;
d = 3 m/s × t and d = 4.8 m/s × (t - 6 s)
At the (common) solution point to the system of equation, we have;
d = 3 m/s × t, d = 4.8 m/s × (t - 6 s)
∴ 3 m/s × t = 4.8 m/s × (t - 6 s)
3·t m/s = 4.8·t m/s - 28.8 m
3·t m/s + 28.8 m = 4.8·t m/s
4.8·t m/s = 3·t m/s + 28.8 m; By transitive property of equality
4.8·t m/s - 3·t m/s = 28.8 m
1.8·t m/s = 28.8 m
t = 28.8 m/(1.8 m/s) = 16 s
t = 16 s.
d = 3 m/s × t = 3 m/s × 16 s = 48 m
d = 48 m
The solution of the system of equation is t = 16 s, d = 48 m
Part C: The possible values of 't' for Boat A are 0 ≤ t ≤ ∞
The possible values of 't' for Boat B are 6 ≤ t ≤ ∞
The timing variable, 't' for the equations of Boat A and Boat B is with reference to the time which Boat A crosses the start line, which is 6 seconds before Boat B, therefore, Boat B crosses the start line when t = 6 seconds
The solution of the system of equation represents that Boat B reaches Boat A at a point 48 meters from the start, 16 seconds from the time Boat A crosses the stat line.
