Answer:
The cars pass next to one another after 25.28 s.
Explanation:
When the cars pass next to one another, the position of both cars is the same relative to the center of the system of reference (marker 0 in this case). Then:
Position of car 1 = position of car 2
The position of an accelerating object moving in a straight line is given by this equation:
x = x0 +v0 t +1/2 a t²
where
x = position at time t
x0 = initial position
v0 = initial speed
t = time
a = acceleration
If the position of car 1 = position of car 2 then:
0 km + 20.0 m/s * t + 1/2 * 0.10 m/s² * t² = 1.2 km - 30.0 m/s * t + 1/2 * 0.30 m/s² * t²
Note that the acceleration of car 2 has to be positive because the car is slowing down and, in consequence, the acceleration has to be opposite to the velocity. The velocity is negative because the direction of car 2 is towards the origin of our system of reference. Let´s continue:
0 km + 20.0 m/s * t + 1/2 * 0.10 m/s² * t² = 1.2 km - 30.0 m/s * t + 1/2 * 0.30 m/s² * t²
1200 m - 50.0 m/s * t + 0.10 m/s² * t² = 0
Solving the quadratic equation:
t = 25.28 s
t = 474. 72 s We discard this value because, if we replace it in the equation of the position of car 2, we will get a position of 20762 m, which is impossible because the position of car 2 can´t be greater than 1200 m.
Then, the cars pass next to one another after 25.28 s
