The cost to print fliers for the opening of an art show varies directly with the number of copies. If it cost $45 to print 750 c
1 answer:
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Answer:
1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.
2) See attached picture.
j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet
j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.
j"(t) represents the acceleration of the car t seconds after the brakes have been hit in
3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.
4)
(see attached picture for graph)
Step-by-step explanation:
1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:
and we set the first derivative equal to zero so we get:
95-18t=0
and solve for t
-18t=-95
t=5.28s
so now we calculate the position of the car after 5.28 seconds, so we get:
so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.
2) For answer 2 I take the second derivative of the function so I get:
j"(t)=-18
and then we graph them. (See attached picture)
j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet
j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.
j"(t) represents the acceleration of the car t seconds after the brakes have been hit in
3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.
4)
(see attached picture for graph)
Let us assume uniform width = x cm wide.
Length of rectangular portrait = 50cm and width of rectangular portrait = 70cm.
Therefore, length of rectangle made by frame = 50 + x+x = (2x+50) cm.
And width of rectangle made by frame = 70+x+x = (2x+70) cm.
We know, the area of rectangular portrait = 50 × 70 = 3500 cm^2.
Total area of the rectangle made by frame would be = (2x+50) * (2x+70)
We know,
Actual area of frame = Area of rectangle made by frame - area of rectangular portrait.
We also given "the area of the frame is the same as the area of the portrait."
We can setup an equation now,
3500 = (2x+50) * (2x+70) - 3500.
Subtracting 3500 from both sides, we get
3500-3500 = (2x+50) * (2x+70) - 3500-3500.
0 = (2x+50) * (2x+70) -7000.
FOIL (2x+50) * (2x+70), we get
0 = 2x*2x +2x*70 + 50*2x +50*70 - 7000.
0 = 4x^2 +140x +100x +3500 -7000.
4x^2 +240x -3500 = 0.
Dividing whole equation by 4, we get
x^2 +60x - 875 =0
Applying quadratic formula , we get
We cant take negative value.
So,
We could take approximately 12 cm.