Ask Question
25 January, 15:25

A red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis. Attimet?0, theredcarisatxr?0andthegreencarisatxg? 220 m. If the red car has a constant velocity of 20 km/h, the cars pass each other at x? 44.5 m, and if it has a constant velocity of 40 km/h, they pass each other at x? 76.6 m. What are (a) the initial velocity and (b) the constant acceleration of the green car?

+5
Answers (1)
  1. 25 January, 18:43
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis. Attimet?0, ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers