Answer:
Let's call the speed of Bug F v_F and the speed of Bug S v_S. Since both bugs started at 0, we can express their positions at any time t as:
Position of Bug F = 12 + v_F * t
Position of Bug S = 8 + v_S * t
To find out when F and S will be 100 units apart, we need to find the time t at which their positions differ by 100 units. In other words, we need to solve the following equation:
|12 + v_F * t - (8 + v_S * t)| = 100
We can simplify this equation by expanding the absolute value and rearranging the terms:
|4 + (v_F - v_S) * t| = 100
Now we can split this equation into two cases:
Case 1: 4 + (v_F - v_S) * t = 100
In this case, we have:
v_F - v_S > 0 (since Bug F is faster)
t = (100 - 4) / (v_F - v_S)
Case 2: 4 + (v_F - v_S) * t = -100
In this case, we have:
v_F - v_S < 0 (since Bug S is faster)
t = (-100 - 4) / (v_F - v_S)
Since we're only interested in positive values of t, we can discard the second case. Therefore, the time at which F and S will be 100 units apart is:
t = (100 - 4) / (v_F - v_S)
t = 96 / (v_F - v_S)
We don't know the values of v_F and v_S, but we can use the fact that Bug F is at 12 and Bug S is at 8, four seconds after they started. This gives us two equations:
12 = 4v_F + 0v_S
8 = 4v_S + 0v_F
Solving these equations for v_F and v_S, we get:
v_F = 3
v_S = 2
Substituting these values into the equation for t, we get:
t = 96 / (3 - 2)
t = 96
Therefore, F and S will be 100 units apart 96 seconds after they start.