Answer:
- 0 ≤ t ≤ 25
- 16.348 seconds
- 310.0 meters
Step-by-step explanation:
a) Since these are production vehicles, we don't expect their top speed to be more than about 70 m/s, so the distance functions probably lose their validity after t = 25. Of course, t < 0 has no meaning in this case, so the suitable domain is about ...
0 ≤ t ≤ 25
Note that the domain for the second car would be 3 ≤ t ≤ 25.
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b) The graph of this system shows the cars will both have driven the same distance after 16.348 seconds.
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c) At that time, the cars will have driven 310.0 meters.
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<em>Non-graphical solution</em>
If you like, you can solve the equation for t:
d1 = d2
1.16t^2 = 1.74(t -3)^2
0 = 0.58t^2 -10.44t +15.66
t = (10.44 +√(10.44^2 -4(0.58)(15.66)))/(2(0.58)) = (10.44+8.524)/1.16
t = 16.348 . . . . time in seconds the cars are at the same distance
That distance is found using either equation for distance:
1.16t^2 = 1.16(16.348^2) = 310.036 . . . meters
