Answer:
The taken is [tex]t_A = 19.0 \ s[/tex]
Explanation:
Frm the question we are told that
The speed of car A is [tex]v_A = 22 \ m/s[/tex]
The speed of car B is [tex]v_B = 29.0 \ m/s[/tex]
The distance of car B from A is [tex]d = 300 \ m[/tex]
The acceleration of car A is [tex]a_A = 2.40 \ m/s^2[/tex]
For A to overtake B
The distance traveled by car B = The distance traveled by car A - 300m
Now the this distance traveled by car B before it is overtaken by A is
[tex]d = v_B * t_A[/tex]
Where [tex]t_B[/tex] is the time taken by car B
Now this can also be represented as using equation of motion as
[tex]d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300[/tex]
Now substituting values
[tex]d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]
Equating the both d
[tex]v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]
substituting values
[tex]29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]
[tex]7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]
[tex]7 t_A =1.2 t_A^2 - 300[/tex]
[tex]1.2 t_A^2 - 7 t_A - 300 = 0[/tex]
Solving this using quadratic formula we have that
[tex]t_A = 19.0 \ s[/tex]