# 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 xr = 0 and the green car is at xg = 216 m. If the red car has a constant velocity of 27.0 km/h ?

Sep 8, 2015

${v}_{0} = - \text{14.56 m/s}$
$a = - {\text{4.96 m/s}}^{2}$

#### Explanation:

FULL QUESTION

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 xr = 0 and the green car is at xg = 216 m.

If the red car has a constant velocity of 27.0 km/h, the cars pass each other at x = 44.2 m. On the other hand, if the red car has a constant velocity of 54.0 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?

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So, you know that the red car starts at $x = 0$ and covers $\text{44.2 m}$ when its velocity is equal to $\text{27.0 km/h}$.

Likewise, the red car starts at $x = 0$ and covers $\text{76.7 m}$ when its velocity is equal to $\text{54.0 km/h}$.

Since the green car's initial velocity and acceleration are the same in both cases, it follows that you can use the time the red car needs to reach those two points to write two equations with two unknowns, the green car's ${v}_{0}$ and $a$.

Now, you need to find the time it takes for both cars to cover their respective distances. To do that, first convert the velocities of the red car to meters per second

27.0color(red)(cancel(color(black)("km")))/color(red)(cancel(color(black)("h"))) * "1000 m"/(1color(red)(cancel(color(black)("km")))) * (1color(red)(cancel(color(black)("h"))))/"3600 s" = "7.5 m/s"

and

54.0color(red)(cancel(color(black)("km")))/color(red)(cancel(color(black)("h"))) * "1000 m"/(1color(red)(cancel(color(black)("km")))) * (1color(red)(cancel(color(black)("h"))))/"3600 s" = "15 m/s"

The time that passes before the cars meet will be

v_1 = d_1/t_1 implies t_1 = d_1/v_1 = (44.2color(red)(cancel(color(black)("m"))))/(7.5color(red)(cancel(color(black)("m")))/"s") = "5.893 s"

and

t_2 = (76.7color(red)(cancel(color(black)("m"))))/(15color(red)(cancel(color(black)("m")))/"s") = "5.113 s"

Now focus on the green car. In the first case, you have that

$44.2 = 216 + {v}_{0} \cdot {t}_{1} + \frac{1}{2} \cdot a \cdot {t}_{1}^{2} \text{ } \textcolor{b l u e}{\left(1\right)}$

The green car starts at $x = \text{216 m}$ and ends up at $x = \text{44.2 m}$, which is where it meets up with the red car.

Likewise, for the second case, you have that

$76.7 = 216 + {v}_{0} \cdot {t}_{2} + \frac{1}{2} \cdot a \cdot {t}_{2}^{2} \text{ } \textcolor{b l u e}{\left(2\right)}$

These will be your two equations with two unknowns.

$\left\{\begin{matrix}- 171.8 = 5.893 \cdot {v}_{0} + a \cdot 17.36 \\ - 139.3 = 5.113 \cdot {v}_{0} + a \cdot 13.07\end{matrix}\right.$

Use the first one to find $a$ as a function of ${v}_{0}$

$a = \frac{- 171.8 - 5.893 \cdot {v}_{0}}{17.36} = - 9.896 - 0.339 \cdot {v}_{0}$

Plug this into the second equation to get

$- 139.3 = 5.113 \cdot {v}_{0} + 13.07 \cdot \left(- 9.896 - 0.339 \cdot {v}_{0}\right)$

$- 139.3 = 5.113 \cdot {v}_{0} - 129.34 - 4.43 \cdot {v}_{0}$

${v}_{0} = \frac{- 139.3 + 129.34}{5.113 - 4.43} = \textcolor{g r e e n}{- \text{14.56 m/s}}$

This means that you have

$a = - 9.896 - 0.339 \cdot \left(- 14.56\right) = \textcolor{g r e e n}{- {\text{4.96 m/s}}^{2}}$

The negative signs you got for the green car's initial velocity and acceleration are consistent with the idea that the green car and the red car are travelling in opposite directions.

If you take the direction of the red car to be positive, then the velocity and acceleration of the green car must come out to be negative.