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 ?

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Sebastian Acosta · Accepted Answer

#v_0 = -"14.56 m/s"#
#a = -"4.96 m/s"""^2# Entire question At time t = 0, the red car is at xr = 0 and the green car is at xg = 216 m. The two cars are moving toward each other in adjacent lanes and parallel to an x axis. The cars pass each other at x = 44.2 m if the red car has a constant velocity of 27.0 km/h, and at x = 76.7 m if the red car has a constant velocity of 54.0 km/h. What are the green car's (a) starting speed and (b) (constant) acceleration? #stackrel("------------------------------------------------------------------------------------------------------------------------")# So, you know that the red car starts at #x=0# and covers #"44.2 m"# when its velocity is equal to #"27.0 km/h"#. Likewise, the red car starts at #x=0# and covers #"76.7 m"# when its velocity is equal to #"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#. The time it takes for each car to travel its respective distances must now be calculated. To do this, convert the red car's velocities 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"# additionally #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 amount of time that elapses before the cars collide is #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"# additionally #t_2 = (76.7color(red)(cancel(color(black)("m"))))/(15color(red)(cancel(color(black)("m")))/"s") = "5.113 s"# Now let's concentrate on the green car. In the initial instance, you have that #44.2 = 216 + v_0 * t_1 + 1/2 * a * t_1^2" "color(blue)((1))# The green car starts at #x="216 m"# and ends up at #x="44.2 m"#, which is where it meets up with the red car. Similarly, in the second instance, you have that #76.7 = 216 + v_0 * t_2 + 1/2 * a * t_2^2" "color(blue)((2))# Your two equations with two unknowns will be these. #{(-171.8 = 5.893 * v_0 + a * 17.36), (-139.3 = 5.113 * v_0 + a * 13.07) :}# Use the first one to find #a# as a function of #v_0# #a = (-171.8 - 5.893 * v_0)/17.36 = -9.896 - 0.339 * v_0# Enter this into the second formula to obtain #-139.3 = 5.113 * v_0 + 13.07 * (-9.896 - 0.339 * v_0)# #-139.3 = 5.113 * v_0 - 129.34 - 4.43 * v_0# #v_0 = (-139.3 + 129.34)/(5.113 - 4.43) = color(green)(-"14.56 m/s")# This indicates that you've #a = -9.896 - 0.339 * (-14.56) = color(green)(-"4.96 m/s"""^2)# The assumption that the green car and the red car are moving in opposite directions is supported by the negative indicators you obtained for the green car's starting velocity and acceleration. The green car's acceleration and velocity must be negative if you assume that the red car's direction is positive.

Isabella Alvarez · Answer

undefined To find the time it takes for the two cars to meet, use the formula:

$ \text{time} = \frac{\text{distance}}{\text{relative velocity}} $

The relative velocity is the sum of the velocities of the two cars since they are moving towards each other. Convert the velocity of the red car from km/h to m/s:

$ 27.0 \, \text{km/h} = \frac{27.0 \times 1000}{3600} \, \text{m/s} $

Then, calculate the relative velocity:

$ \text{relative velocity} = \text{velocity of red car} + \text{velocity of green car} $

Finally, plug the values into the formula to find the time it takes for the two cars to meet.