Answer:
the correct solution is 13 s
Explanation:
This is a kinematic problem, let's use accelerated rectilinear motion relationships.
For the first car it has an accelerometer of 2.0 m/s²
x = v₀₁ t + ½ a₁ t²
The second car leaves the same point, but 4.0 seconds later
x = v₀₂ (t-4) + ½ a₂ (t-4)²
With this form we use the same time for both cars.
The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position
x = ½ a₁ t²
x = ½ a₂ (t-4)²
Let's solve
a₁ t² = a₂ (t-4)²
a₁/a₂ t² = t² -2 4 t + 16
t² (1- 2.0 / 4.0) - 8 t +16
t² 0.5 - 8 t +16 = 0
t² -16 t + 32 = 0
Let's solve the second degree equation
t = [16 ±√( 16² - 4 32)] / 2
t = ½ (16 ± 11,3)
Solutions
t1 = 13.66 s
t2 = 2.34 s
These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s
the correct solution is 13 s, if you have to select one the nearest 12s
