A red car and a green car, identical except for the color, move toward each othe
ID: 1544556 • Letter: A
Question
A red car and a green car, identical except for the color, 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 = 219 m. If the red car has a constant velocity of +20 km/h, the cars pass each other at x = 43.8 m, and if it has a constant velocity of +40 km/h, they pass each other at x = 76.3 m. (a) What is the initial velocity of the green car? (Indicate direction with the sign of your answer.) ______ m/s (b) What is the acceleration of the green car? (Indicate direction with the sign of your answer.) ______ m/s2 HINT: Constant-acceleration equations. There are two pairs of data for a red-car speed and a passing point. For each pair, set up two simultaneous equations, one for each car. At the moment of passing, the cars have the same coordinate. Use that fact to merge the two simultaneous equations.
Explanation / Answer
Here for red car a=0
Case (i)
Constant velocity V1=20 Km/hr=50/9 m/s
We know velocity=distance/time taken
For x=43.8, V1=50/9
Time,t1=Dist/Velocity
t1 = 43.8/50/9=7.9s
Case (ii)
Constant velocity V2=40 Km/hr=100/9 m/s
We know velocity=distance/time taken
For x=76.3, V2=100/9
Time,t2=Dist/Velocity
t2 = 76.3/100/9=6.9s
Now we must solve two simultaneous equations for the given car
43.8-219=V0 t1 +at12/2 ------ 1
76.3-219=V0t2+at22/2 ------ 2
V0 t1 +at12/2 =-175.2 ------ 1
V0t2+at22/2 =-142.7 ------ 2
Eqn1 x t1 V0t1t2+at12t2=-175.2t2 ------------3
Eqn1 x t2 V0t1t2+at22t1=-142.7t1 ------------4
(-) (-) =(+)
at12t2- at22t1=-175.2t2+142.7t1
215.3a-188.05a = -1208.88+1127.33
27.55a=-81.55
Acceleration,a=-81.55/27.55=-2.99 m/s2
V0 t1 +at12/2 =-175.2
7.9Vo-93.30=-175.2
Initial velocity,V0=-10.366 m/s
Here for red car a=0
Case (i)
Constant velocity V1=20 Km/hr=50/9 m/s
We know velocity=distance/time taken
For x=43.8, V1=50/9
Time,t1=Dist/Velocity
t1 = 43.8/50/9=7.9s
Case (ii)
Constant velocity V2=40 Km/hr=100/9 m/s
We know velocity=distance/time taken
For x=76.3, V2=100/9
Time,t2=Dist/Velocity
t2 = 76.3/100/9=6.9s
Now we must solve two simultaneous equations for the given car
43.8-219=V0 t1 +at12/2 ------ 1
76.3-219=V0t2+at22/2 ------ 2
V0 t1 +at12/2 =-175.2 ------ 1
V0t2+at22/2 =-142.7 ------ 2
Eqn1 x t1 V0t1t2+at12t2=-175.2t2 ------------3
Eqn1 x t2 V0t1t2+at22t1=-142.7t1 ------------4
(-) (-) =(+)
at12t2- at22t1=-175.2t2+142.7t1
215.3a-188.05a = -1208.88+1127.33
27.55a=-81.55
Acceleration,a=-81.55/27.55=-2.99 m/s2
V0 t1 +at12/2 =-175.2
7.9Vo-93.30=-175.2
Initial velocity,V0=-10.366 m/s
Here for red car a=0
Case (i)
Constant velocity V1=20 Km/hr=50/9 m/s
We know velocity=distance/time taken
For x=43.8, V1=50/9
Time,t1=Dist/Velocity
t1 = 43.8/50/9=7.9s
Case (ii)
Constant velocity V2=40 Km/hr=100/9 m/s
We know velocity=distance/time taken
For x=76.3, V2=100/9
Time,t2=Dist/Velocity
t2 = 76.3/100/9=6.9s
Now we must solve two simultaneous equations for the given car
43.8-219=V0 t1 +at12/2 ------ 1
76.3-219=V0t2+at22/2 ------ 2
V0 t1 +at12/2 =-175.2 ------ 1
V0t2+at22/2 =-142.7 ------ 2
Eqn1 x t1 V0t1t2+at12t2=-175.2t2 ------------3
Eqn1 x t2 V0t1t2+at22t1=-142.7t1 ------------4
(-) (-) =(+)
at12t2- at22t1=-175.2t2+142.7t1
215.3a-188.05a = -1208.88+1127.33
27.55a=-81.55
Acceleration,a=-81.55/27.55=-2.99 m/s2
V0 t1 +at12/2 =-175.2
7.9Vo-93.30=-175.2
Initial velocity,V0=-10.366 m/s
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