To find where the truck and the car meet, we need to determine the time it takes for both vehicles to meet each other.
Let's start by finding the time it takes for the car to reach the meeting point. We can use the following kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the car comes to a stop)
u = initial velocity of the car (-26 m/s, since it's traveling south)
a = acceleration of the car (-3.5 m/s^2, as it slows down)
s = displacement of the car (unknown)
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Plugging in the given values, we get:
s = (0^2 - (-26)^2) / (2 * -3.5)
s = (0 - 676) / -7
s = 97.14 m
Using the same method for the truck, we can find the displacement (distance traveled by the truck) in the time it takes for the car to reach the meeting point. Since the truck is moving at 19 m/s to the north, the initial velocity of the truck is 19 m/s. The truck will continue to travel at a constant speed until it reaches the meeting point.
Let's calculate the time it takes for the car to reach the meeting point:
s = ut + (1/2)at^2
Where:
s = displacement of the car (97.14 m, as calculated earlier)
u = initial velocity of the car (-26 m/s)
a = acceleration of the car (-3.5 m/s^2)
t = time taken to reach the meeting point (unknown)
Rearranging the equation, we have:
0 = -26t + (1/2)(-3.5)t^2
This is a quadratic equation. To solve for t, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values from the equation, we get:
t = (-(-26) ± √((-26)^2 - 4 * (-3.5) * 0)) / (2 * (-3.5))
t = (26 ± √(676)) / (-7)
t = (26 ± 26) / (-7)
This gives us two solutions for t:
t1 = 4 s (ignoring the negative solution)
Now, we can find the displacement of the truck in the same amount of time. The truck is moving at a constant speed of 19 m/s to the north, so its displacement can be calculated as:
s = ut
s = 19 m/s * 4 s
s = 76 m
Therefore, the truck and the car meet 76 meters north of the initial positions of both vehicles.