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natka813 [3]
3 years ago
5

The state of illinois cycle rider safety program requires motorcycle riders to be able to brake from 30 mph (44 feet/second) to

zero in 45 feet. what constant deceleration is required to do that?
Physics
1 answer:
-Dominant- [34]3 years ago
6 0
<span>21.5 ft/s^2 The formula for distance traveled under constant acceleration is d = 0.5 AT^2 where d = distance A = acceleration T = time Solving for A, gives d = 0.5 AT^2 2d = AT^2 (1) 2d/T^2 = A The formula for velocity under constant acceleration is V = AT Solving for A, gives V = AT (2) V/T = A Now setting equations (1) and (2) above equal to each other and solving for T: V/T = 2d/T^2 TV = 2d (3) T = 2d/V Now substitute equation (3) into (2) above. V/T = A V/(2d/V) = A V * V/2d = A V^2/2d = A Plug in values and calculate. V^2/2d = A (44 ft/s)^2/(2*45 ft) = A (1936 ft^2/s^2)/(90 ft) = A 21.51111111 ft/s^2 = A Rounding to 3 significant figures gives 21.5 ft/s^2</span>
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Answer:

the rocks are away from cracks in the crust

Explanation:

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2 years ago
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A bowling ball is launched off the top of a 240-foot tall building. The height of the bowling ball above the ground t seconds af
natima [27]

Answer:

When the ball hits the ground, its velocity is -128 ft/s.

Explanation:

Hi there!

First, let's find the time it takes the ball to reach the ground (the value of t for which s(t) = 0):

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0 = -16t² + 32t + 240

Solving the quadratic equation with the quadratic formula:

t = 5.0 s (the other solution of the equation is rejected because it is negative).

Now, we have to find the velocity of the ball at t = 5.0 s.

The velocity of the ball is the change of height over time (the derivative of s(t)):

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at t = 5.0 s:

s'(5.0) = -32(5.0) + 32 = -128 ft/s

When the ball hits the ground, its velocity is -128 ft/s.

4 0
3 years ago
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