In the figure here, a red car and a green car move toward each other in adjacent

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, = 0 and the green car is at xg 211 m. If the red car has a constant velocity of 21.0 km/h, the cars pass each other at x - 43.8 m. On the other hand, if the red car has a constant velocity of 42.0 km/h, they pass each other at x 76.3 m. What are (a) the initial velocity and (b) the (constant) acceleration of the green car? Include the signs. Green xy car ed car (a) Number Unit km/h nit m/s2 (b) Number

Explanation / Answer

Solution:

Let us go to the basics first.

Vr = 21km/h * 1m/s / 3.6km/h = 5.833 m/s:
Sr =43.8 m = 5.833 m/s * t
this tells us t = 43.8 /5.833 s = 7.5 s
Sg = 43.8 m = 211m + Vg * t + ½at²
Plug in for t:
43.8 = 211 + 7.5Vg + 28.125a
-7.5Vg = 167.2 + 28.125a
Vg = -22.293 - 3.75a for later

Vr = 42km/h * 1m/s / 3.6km/h = 11.67 m/s:
Sr = 76.3m = 11.67m/s * t
t = 6.538 s
Sg = 76.3m = 211m + Vg * t + ½at²
Sub for t and Vg:
76.3 = 211 + (-22.293 - 3.75a)( 6.538) + ½a(6.538)²
76.3 = 211 - 145.751 - 24.5175a + 21.372a
11.051 = -3.1455a
a = -3.513 m/s² acceleration (Answer b)
Vg = -22.293 - 3.75 (-3.513) = -9.12 m/s = - 32.832 km/h initial velocity (Answer a)

Thanks!!!

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