Question

Plz help

Two rowing teams are having a race. Boat A had a head start, crossing the start line first and traveling at a speed of 3 meters per second. Boat B waited and then crossed the start line 6 seconds after Boat A. The speed of Boat B was 4.8 meters per second.

Part A:
Write two equations, one for Boat A and one for Boat B. In both equations, let t equal the time in seconds since Boat A crossed the start line and d equal the distance traveled in meters.

Part B:
Solve the system of equations from Part A. Show your work.

Part C:
Determine the possible values of t for each of the equations from Part A and give a reason for your answers. Then explain what the solution to the system of equations from Part B represents in the context of the problem.

Answer 1

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

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.