Answer:
13.7 s
Explanation:
The position at time t of car A can be written as follows:
[tex]x_A (t) = \frac{1}{2}a_At^2[/tex]
where
[tex]a_A = 11 m/s^2[/tex] is the acceleration of car A
The position of car B instead can be written as
[tex]x_B(t) = d+\frac{1}{2}a_B t^2[/tex]
where
[tex]a_B = -4 m/s^2[/tex] is the acceleration of car B
d = 1400 m is the initial separation between the cars
The two cars meet when
[tex]x_A = x_B[/tex]
Using the two equations above,
[tex]\frac{1}{2}a_A t^2 = d + \frac{1}{2}a_B t^2\\\frac{1}{2}t^2 (a_A - a_B) = d\\t=\sqrt{\frac{2d}{a_A-a_B}}=\sqrt{\frac{2(1400)}{11-(-4)}}=13.7 s[/tex]
