Question

PLEASE HELP!! A car manufacturer does performance tests on its cars. During one test, a car starts from rest, and accelerates at a constant rate for 20 seconds. another car starts from rest three seconds later, and accelerates at a faster constant rate. The equation that models the distance (d) in metres the first cars equation is d=1.16t^2, where t is time, in seconds, after the car starts. The equation for the second car is: d=1.74(t-3)^2. a) in context, what is a suitable domain for the graph of the system? b) at what time will both cars have driven the same distance? c) how far will they have driven at this time?

Answer 1

• 0 ≤ t ≤ 25
• 16.348 seconds
• 310.0 meters

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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Non-graphical solution

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