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4 years ago

15

A car advertisement claims their car can go from a stopped position to moving 60 miles per hour in 5 seconds. The advertisement

is describing the car's acceleration. speed. velocity. direction.

1 answer:

ASHA 777 [7]4 years ago

4 0

The advertisement is deciribing the car’s acceleration, as it gains speed when moving from 0-60.

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61 degrees, I just did the test.

Explanation:

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In order to answer this exercise you need to use the formulas

S = Vo*t + (1/2)*a*t^2

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A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

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Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

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$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

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$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

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$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

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