Answer:
car A reaches and immediately overtakes the car B at 22.56 s.
Explanation:
After car A accelerate at 1.8 m/s2, it travels a distance x(A) and car B will have travels a distance x(B), let's recall that the initial distance between them is 300 m, so we have:
[tex]x_{A}=300+x_{B}[/tex]
Now, we can rewrite this equation in terms of speed and time
[tex]V_{iA}t+\frac{1}{2}at^{2}=300+V_{iB}t[/tex]
Where:
V(iA) is the initial speed of car A
V(iB) is the initial speed of car B
t is the time when car A reaches the car B
a is the acceleration
[tex]18t+\frac{1}{2}1.8t^{2}=300+25t[/tex]
[tex]0.9t^{2}-7t-300=0[/tex]
Solving this quadratic equation for t, and taking just the positive value, we will have:
t=22.56 s
Therefore, car A reaches and immediately overtakes the car B at 22.56 s.
I hope it helps you!
