SOLUTION: The relation d=0.0052s^2 + 0.13s models the stopping distance, d, in metres, of a car travelling at a speed of s, in kilometres per hour, when the driver breaks hard. At what speed
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Question 1007977: The relation d=0.0052s^2 + 0.13s models the stopping distance, d, in metres, of a car travelling at a speed of s, in kilometres per hour, when the driver breaks hard. At what speed was a car travelling if its stopping distance is 20m? Round to the nearest tenth.
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The relation d=0.0052s^2 + 0.13s models the stopping distance, d, in metres, of a car travelling at a speed of s, in kilometres per hour, when the driver breaks hard. At what speed was a car travelling if its stopping distance is 20m? Round to the nearest tenth.
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Solve::
0.0052s^2 + 0.13s - 20 = 0
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s = [-0.13 +- sqrt(0.13^2 - 4*0.0052*-20)]/(2*0.0052)
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s = [-0.13 +- sqrt(0.4329)]/(0.0104)
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s = [-0.13 +- 0.6580]/0.0104
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Positive solution::
s = [-0.13+0.6580]/0.0104
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s = 50.8 km/hr
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Cheers,
Stan H.
Answer by MathLover1(20849) (Show Source): You can put this solution on YOUR website!
we are given , and
the equation becomes
and we will use the following quadratic formula to solve it for ,
solutions:
or
...disregard negative solution
and your solution is:
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