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. At time t = 0, the red car is at xr = 0 and the green car is at xg = 220 m. If the red car has a constant velocity of +20 km/h, the cars pass each other at x = 44.8 m, and if it has a constant velocity of +40 km/h, they pass each other at x = 77.7 m.

(a) What is the initial velocity of the green car? (Indicate direction with the sign of your answer.)
______ m/s

(b) What is the acceleration of the green car? (Indicate direction with the sign of your answer.)
______ m/s2

Answer 1

(a) -11.356 m/s

(b) -2.572 m/s²

Explanation:

Let a and v represent the green car acceleration in m/s² and velocity in m/s, respectively. Then the position of the green car in meters at time t is g(t):

g(t) = 1/2at² +vt +220

In the first scenario, the cars pass at g(t) = 44.8 for t = 44.8/(20/3.6) = 8.064.

In the second scenario, the cars pass at g(t) = 77.7 for t = 77.7/(40/3.6) = 13.986.

This lets us write two equations in "a" and "v":

32.514048a +8.064v +220 = 44.8

24.4510245a +6.993v +220 = 77.7

There are numerous ways to solve this pair of linear equations. For our purpose, a graphical solution is sufficiently accurate. It tells us ...

a = -2.572 m/s²

v = -11.356 m/s

(a) The initial velocity of the green car is -11.356 m/s.

(b) The acceleration of the green car is -2.572 m/s².

_____

The graph has meters on the vertical axis and seconds on the horizontal axis. The green curve is the position of the green car as a function of time. The purple and black lines are the position of the red car at 20 kph and 40 kph, respectively.

The red and blue lines are the equations above, with x=a, y=v. For those equations, the x-axis is m/s², and the y-axis is m/s.