Two cars emerge side by side from a tunnel. car a is traveling with a speed of 60km/h and has an acceleration of 40km/h/min. car b has a speed of 40km/h and has an acceleration of 60 km/h/min. which car is passing the other as they come out of the tunnel?

To solve this problem, we have to make use of the formula:

d = v0 t + 0.5 a t^2

What we can do is to assume a time, t = 1 h.

For Car A:

d = 60 * 1 + 0.5 (40 * 60) * 1^2

d = 1,260 km

For Car B:

d = 40 * 1 + 0.5 (60 * 60) * 1^2

d = 1,840 km

Since Car B travelled the larger distance, so it is the Car which passes the other as they come out.

Car B

Car A will pass Car B first when they come out of the tunnel.

       To solve this question, we just need to use simple logic.

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when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

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