Question

In the figure here, a red car and a green car move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at x_r = 0 and the green car is at x_g = 213 m. If the red car has a constant velocity of 22.0 km/h, the cars pass each other at x = 43.6 m. On the other hand, if the red car has a constant velocity of 44.0 km/h, they pass each other at x = 76.7 m. What are (a) the initial velocity and (b) the (constant) acceleration of the green car? Include the signs.

Answer 1

Let's first convert the velocity of the red car in both cases from km/h to m/s:

$$v_1=22.0\ \dfrac{km}{h}\cdot\dfrac{1000\ m}{1\ km}\cdot\dfrac{1\ h}{3600\ s}=6.11\ \dfrac{m}{s},$$ $$v_2=44.0\ \dfrac{km}{h}\cdot\dfrac{1000\ m}{1\ km}\cdot\dfrac{1\ h}{3600\ s}=12.22\ \dfrac{m}{s}.$$

Then, we can find the time when the two cars  pass each other for both cases:

$$t_1=\dfrac{x_1}{v_1}=\dfrac{43.6\ m}{6.11\ \dfrac{m}{s}}=7.1\ s,$$ $$t_2=\dfrac{x_2}{v_2}=\dfrac{76.7\ m}{12.22\ \dfrac{m}{s}}=6.3\ s.$$

Let's write the equations of motion of the green car in both cases:

$$x_{1f}-x_{1i}=v_0t_1+\dfrac{1}{2}at_1^2,$$ $$x_{2f}-x_{2i}=v_0t_2+\dfrac{1}{2}at_2^2.$$

Then, we get:

$$43.6-213=7.1v_0+\dfrac{1}{2}a(7.1)^2,$$ $$76.7-213=6.3v_0+\dfrac{1}{2}a(6.3)^2.$$

After simplification, we get:

$$7.1v_0+25.2a=-169.4,$$ $$6.3v_0+19.84a=-136.3.$$

Let's express $a$ from the first equation in terms of $v_0$:

$$a=\dfrac{-169.4-7.1v_0}{25.2}.$$

Substituting $a$ into the second equation, we get:

$$6.3v_0+\dfrac{19.84}{25.2}(-169.4-7.1v_0)=-136.3.$$

From this equation, we can find $v_0$:

$$17.9v_0=-73.96,$$ $$v_0=\dfrac{-73.96}{17.9}=-4.13\ \dfrac{m}{s}.$$

Then, substituting $v_0$ into the expression for $a$ we get:

$$a=\dfrac{-169.4-7.1\cdot(-4.13)}{25.2}=-5.56\ \dfrac{m}{s^2}.$$