SOLUTION: Hi tutor, I am stuck in the question written below. Can you please tell to me on how to solve it. The question: The relation d=0.0052s^2 + 0.13s models the stopping distance,

Question 737588: Hi tutor, I am stuck in the question written below. Can you please tell to me on how to solve it.
The question:
The relation d=0.0052s^2 + 0.13s models the stopping distance,d, in meters, of a car travelling at a speed of s, in kilometers per hour, when the driver brakes hard. At what speed was a car travelling if its stopping distance is 20m? Round to the nearest tenth.
Can you please show me how to solve this question step by step. Thank you.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
we are given d=0.0052s^2 + 0.13s, and d=20meters
the equation becomes
0.0052s^2 + 0.13s - 20 = 0 and we will use the following quadratic formula to solve it for s,
s = (-0.13 + sqrt(0.13^2 - 4*0.0052*-20)) / (2*0.0052) = 50.76
s = (-0.13 - sqrt(0.13^2 - 4*0.0052*-20)) / (2*0.0052) = -75.76
and
s= 50.8 kilometers per hour
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