Note the expression y - x^2, Which statement is most consistent with this expres

Question

Note the expression y - x^2, Which statement is most consistent with this expression? If x doubles, then y quadruples if y doubles, then x quadruples if x triples, then y doubles An object is thrown vertically and has an upward velocity of 25 m/s when it reaches one-fourth of its maximum height above its launch point. What is the maximum height of the object? 50 m 76.37 m 32.46 m 42.52 m 11.68 A ball is shot with a velocity of 36 m/s at an angle of 30 degree from the top of a 20 m call building. What is the horizontal distance the ball travels when it hits the ground? 142 m 72 m 240 m 96 m 118 m A cart is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.5 m/s^2., What is the magnitude of the cart's displacement during the first 6.0 s of its motion? 55 m 10 m 100 m 75 m 66 m At the top of a cliff 100 m high, Raoul throws a rock upward with velocity 20 m's. How much later should he drop a second rock from rest so both rocks arrive simultaneously at the bottom of the cliff? 5.05s 3.76s 1.78s 2.48s 0.81 s An airplane starts from rest and accelerates at 12.1 m/s^2. What is its speed at the end of a 500 m runway? 98 ms The position of a particle moving along the x-axis is given by x - 30 t^2 - 30 t^4, where x is in meters and t is in seconds. What is the position of the particle when it achieves its maximum speed in the positive x-direction? 5.6 m 9.8 m 4.2 m 27 m 3 m When a drag strip vehicle reaches a velocity of 60 m's. it begins a negative acceleration by releasing a drag chute and applying its brakes. While reducing its velocity back to zero, its acceleration along a straight line path b a constant -9.5 m/s^2., What displacement docs it undergo during this deceleration period? 40 m 320 m 77 m 240 m 189 A car slows down from 3 speed of 31 m/s to a speed of 12 m/s over a distance of 380 m. How long does this take, assuming constant acceleration?

Explanation / Answer

a)
y=x^2

let say, x'=2x
then
y' = (2x)^2
= 4*x^2
= 4*y

so y quadruples

Answer: A

b)
Let maximum height be h m
at max height, vf = 0
vi = v m/s
a = -9.8 m/s^2
use:
vf^2 = vi^2 + 2*a*d
0 = v^2 - 2*9.8*h
v = sqrt (2*9.8*h)

now, for h/4 m
vf = 25 m/s
vi = v m/s
a = -9.8 m/s^2
d = h/4 m

use:
vf^2 = vi^2 + 2*a*d
25^2 = (2*9.8*h) - 2*9.8*h/4
25^2 = 3*2*9.8*h/4
h = 42.52 m
Answer: D

i am allowed to answer only 1 question at a time

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