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?
1.
given taht, y = x^2
lry x\'= 2x, then the new value of y is,
y\' =(2x)^2 = 4x^2 = 4y
therefore, option A is correct.
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2.
kinematic equation 1: 25*25 = vyi^2-4.9ymax+19.6
kinematic equation 2: -0 = vyi^2-19.6ymax+19.6
then the value of ymax is,
ymax= 42.517 m
but, ymax/4 = 10.63 m, therefore, the velocity is,
25*25 = vyi^2-19.6(10.63)
vyi = 28.8675 m/s
maximum height, hmax = vyi^2/2g = 42.52 m.
option D is correct answer.
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3.
time taken to reach the ground.
y = v0yt - 0.5gt^2
-20 = (36*sin30)*t - 0.5*9.8*t^2
after solve, we get, t = 4.567 s
hence, the horizontal distance is,
R = vxt = v0cos30*t = 36*cos30*4.567 = 142 m
option A is correct
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4.
dispalcement is,
s = ut + 0.5at^2
= 5*6 + 0.5*2.5*6*6
= 75 m
Option D is correct.
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