a) The initial velocity of the green car is -13 m/s
b) The acceleration of the green car is - 2.25 m/s²
Explanation:
The equation for the position of objects moving in a straight line with constant acceleration is as follows:
x = x0 + v0·t + 1/2·a·t²
where:
x = position
x0 = initial position
v0 = initial velocity
a = acceleration
t = time
If the velocity is constant, then a = 0 and x = x0 + v·t
a)The initial position of the red and green car is 0 m and 220 m respectively. We know that at 44.5 m the cars pass each other if the red car has a constant velocity of 20 km/h. So let´s find how much time it takes the cars to pass each other in this case:
The position of the red car is:
x = x0 + v·t
then:
0.0445 km = 0 km + 20 km/h · t
t = 0.0445 km/ 20 km/h = 8.0 s
We also know that if the red car has a velocity of 40 km/h, both cars pass each other at 76.6 m. So let´s find the time it takes the cars to reach that position using the equation for the red car:
0.0766 km = 0 km + 40 km/h · t
t = 0.0766 km / 40 km/h = 6.9 s
The position of the green car at t= 6.9 s and t = 8.0 s must be the same as the red car because both cars pass each other at those times.
Then, for the green car:
x = x0 + v0·t + 1/2·a·t²
0.0445 km = 0.220 km + v0 · 8.0 s + 1/2·a· (8.0 s)²
and
0.0766 km = 0.220 km + v0 · 6.9 s + 1/2·a· (6.9 s)²
Now we have a system of two equations with two unknowns.
Solving for "a" in the first equation
0.0445 km - 0.220 km - v0 · 8.0 s = 32 s²·a
(-0.176 km - v0 · 8.0 s) / 32 s² = a
Replacing a = (-0.176 km - v0 · 8.0 s) / 32 s² in the second equation and solving for v0:
0.0766 km = 0.220 km + v0 · 6.9 s + 1/2·((-0.176 km - v0 · 8.0 s)/32 s²)·(6.9 s)²
-0.143 km = v0 · 6.9 s - 0.74(0.176 km + v0 · 8.0 s)
-0.143 km = v0 · 6.9 s - 0.130 km - v0 · 5.9 s
-0.143 km + 0.130 km = v0 · 6.9 s - v0 · 5.9 s
-0.013 km = 1 s · v0
v0 = -13 m/s
b) The acceleration of the green car is:
a = (-0.176 km - v0 · 8.0 s) / 32 s²
a = (-0.176 km - (-0.013 km/s) · 8.0 s) / 32 s² = -2.25 m/s²
