5. You need to find the integral to find the equation to give distance. The car is6x. The anti for bike is .5x .5x + 6x = 7 miles ( the distance separating) X = 1.077 so the car and the bike would have collide in approximately 1 sec. The position of the car at this point was at 6 * 1.077 = 6.4625 The position of the bike at this point was .5 * 1.077 = .5385
6. The car in one second would have only been going 6 miles a second since in 1 seconds that’s the only amount of speed that could have been picked up but the position would have been 6 miles away from the original stopping point and the car to travel that far at the steady increase would have been going 36 miles an hour and therefore the data was recorded wrong and can be proven wrong and the driver is innocent .
7.
8. To find the maximum of the hill find the 0, and undefined points of the hill. -.1x^2 -.15x + 4.5 the derivative is -.2x - .15 There are no undefined points -.2x - .15 = 0 x=-.75 -.1(-.15) ^2 - .15(-.75) + 4.5 = 4.55625 Find the slope of the point the car is at. Take the derivative of the original equation -.2x-.15 plug in the point the car is at = 3.5 The slope of the line is m = .55 Y= mx + b 3.8= .55(-3.5) + b b= 5.725 y= .55x + 5.725
10. Draw the triangle created by y= 4.55625 and y=.55x + 5.725 and the horizontal line x = -3.5 Find the angel created at the intersection of the vertical line and the slope of the car tan x = .75625/1.375 Angle = 28.81 The angle of the car would be able to see the bicyclist and should have been able to stop (3.5, 4.55625) (3.5, 3.8) (-2.125, 4.55625)
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