Question:
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
Answer:
A. Equations
[tex]d = 3t[/tex]
[tex]d = 4.8t - 28.8[/tex]
b.
[tex]t = 16[/tex]
[tex]T = 10[/tex]
[tex]d = 48km[/tex]
Step-by-step explanation:
For Boat A:
[tex]Speed = 3m/s[/tex]
[tex]Time = t[/tex]
For Boat B:
[tex]Speed = 4.8m/s[/tex]
Boat B waited 6 seconds later. So, the time is:
[tex]T = t - 6[/tex]
Solving (a): Equations
Speed is calculated as:
[tex]Speed = \frac{Distance}{Time}[/tex]
For Boat A:
[tex]3 = \frac{d}{t}[/tex]
Make d the subject
[tex]d = 3t[/tex]
For Boat B:
[tex]4.8 = \frac{d}{T}[/tex]
Make d the subject
[tex]d = 4.8T[/tex]
Substitute t - 6 for T
[tex]d = 4.8(t - 6)[/tex]
[tex]d = 4.8t - 28.8[/tex]
Hence, the equations are:
[tex]d = 3t[/tex]
[tex]d = 4.8t - 28.8[/tex]
Solving (b): The value of d
[tex]d = 3t[/tex]
[tex]d = 4.8t - 28.8[/tex]
Substitute 3t for d in the second equation
[tex]3t = 4.8t - 28.8[/tex]
Collect like terms
[tex]3t - 4.8t = - 28.8[/tex]
[tex]- 1.8t = - 28.8[/tex]
[tex]1.8t = 28.8[/tex]
Solve for t
[tex]t = 28.8/1.8[/tex]
[tex]t = 16[/tex]
Recall that:
[tex]T = t - 6[/tex]
[tex]T = 16 - 6 = 10[/tex]
Next, calculate the distance:
[tex]d = 3t[/tex]
[tex]d = 3 * 16[/tex]
[tex]d = 48km[/tex]
So:
Boat A travels for 16 seconds
Boat B travels for 10 seconds
The distance is 48km
