a bird flying over water drops a crab from a height of 10 feet. the distance the crab is from the water as…

a bird flying over water drops a crab from a height of 10 feet. the distance the crab is from the water as it falls can be represented by d in the equation d = -t² + 10, where t is time in seconds. to catch the crab as it falls, a different bird flies along a path represented by the equation d = 5t + 5. in how many seconds will the second bird catch the crab before it hits the water?

a bird flying over water drops a crab from a height of 10 feet. the distance the crab is from the water as it falls can be represented by d in the equation d = -t² + 10, where t is time in seconds. to catch the crab as it falls, a different bird flies along a path represented by the equation d = 5t + 5. in how many seconds will the second bird catch the crab before it hits the water?

Answer

Explanation:

Step1: Set the two equations equal

Since the bird catches the crab when their distances (d) are equal. The crab's distance is (d=-t^{2}+10) and the bird's distance is (d = 5t+5). So, we set (-t^{2}+10=5t + 5).

Step2: Rearrange to standard quadratic form

Move all terms to one side: (t^{2}+5t - 5=0). The quadratic formula for (ax^{2}+bx + c=0) is (t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}). Here (a = 1), (b=5), (c=-5).

Step3: Calculate the discriminant

(\Delta=b^{2}-4ac=(5)^{2}-4\times1\times(-5)=25 + 20=45).

Step4: Apply the quadratic formula

(t=\frac{-5\pm\sqrt{45}}{2}=\frac{-5\pm3\sqrt{5}}{2}). We have two solutions (t=\frac{-5 + 3\sqrt{5}}{2}\approx\frac{-5+6.708}{2}\approx0.85) and (t=\frac{-5 - 3\sqrt{5}}{2}\approx\frac{-5 - 6.708}{2}\approx-5.85). Since time (t\geq0), we reject the negative solution.

Answer:

(0.85) seconds