In the figure here, a red car and a green car move toward each other in adjacent
ID: 1400998 • Letter: I
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 xr = 0 and the green car is at xg = 222 m. If the red car has a constant velocity of 26.0 km/h, the cars pass each other at x = 43.7 m. On the other hand, if the red car has a constant velocity of 52.0 km/h, they pass each other at x = 76.4 m. What is the initial velocity
Answer has to be in km/hr. Please show work as I'd like to learn how to solve the problem. Thank you so much.
Explanation / Answer
At time t = 0,
the red car is at xr = 0 and the green car is at xg = 222 m.
distance for red car, x1 = v1 t1
or t1 = x1 / v1 { eq.1 }
where, v1 = velocity of red car = (26 km / hr) x (1000 m / 3600 sec) = 7.22 m/s
x1 = 43.7 m
inserting the values in eq.1,
t1 = (43.7 m) / (7.22 m/s)
t1 = 6.05 sec
another distance for red car, x2 = v2 t2
or t2 = x2 / v2 { eq.2 }
where, v2 = constant velocity of red car = 52 km/hr = (52 km/hr) x (1000 m / 3600 sec) = 14.4 m/s
x2 = 76.4 m
inserting the values in eq.2,
t2 = (76.4 m) / (14.4 m/s)
t2 = 5.3 sec
Now, using equations of motion 2 for green cars -
x1 = xg + v0 t1 + (1/2) a t12 { eq.3 }
x2 = xg + v0 t2 + (1/2) a t22 { eq.4 }
solving two above eq.
(43.7 m) - (222 m) = (6.05 sec) v0 + (18.3) a
(6.05) v0 + (18.3) a = -178.3 { eq.5 }
inserting the values in eq.4,
(76.4 m) - (222 m) = (5.3 sec) v0 + (14) a
(5.3) v0 + (14) a = -145.6 { eq.6 }
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(6.05) v0 + (18.3) a = -178.3 x 14
(5.3) v0 + (14) a = -145.6 x 18.3
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(84.7) v0 + (256.2) a = -2496.2
(96.9) v0 + (256.2) a = -2664.4
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-12.2 v0 = 168.2
v0 = -13.7 m/s
velocity change into km/hr :
v0 = - 49.3 km/hr
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