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4 years ago

A car is driving 25.5 m/s when it hits

Physics

1 answer:

IceJOKER [234]4 years ago

6 0

Answer:

it takes the car 4.362 seconds to cover the distance of 88.4 m.

Explanation:

The distance the car covers is given by the function

$x(t) = vt-\dfrac{1}{2}at^2$ ,

where $v = 25.5m/s$ , and $a = 2.40m/s$ , putting these in we get:

$x(t) = 25.5t-\dfrac{1}{2}(2.4)t^2\\\\x(t) = 25.5t-1.2t^2$

Now, when the car has moved to 88.4m, $x(t) = 88.4$ , or

$x(t) = 25.5t-1.2t^2 = 88.4$

which is a quadratic equation with solutions

$t = 4.362s,\\t = 16.887s$

We take the first solution $t = 4.362s$ , since at that time the car is still moving right and decelerating. The second solution $t =16.887$ describes the situation where the car has stopped decelerating and is now moving leftwards because the decelerating is leftwards, which is utterly wrong because we know that cars do not start moving backwards after the brakes have stopped them!

Thus, it takes the car 4.362 seconds to cover the distance of 88.4 m.

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Sarah's acceleration is $-0.49 m/s^2$

Explanation:

The force of kinetic friction acting on Sarah has a magnitude which is given by:

$F_f = \mu mg$

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Moreover, according to Newton's second law of motion, we know that the net force on Sarah is equal to its mass times its acceleration:

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$F=-\mu mg = ma$

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we find:

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3 years ago

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ASSUMPTION:

we neglect the change in density across 100m vertically

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we equate the pressure for mercury and air

AIR(pgh) = Mercury(pgh)

0.8 * g * 100 = 13600 * g * H

H = 0.00058m = 0.58mm of Mercury

At even farther heights, the air density further drops

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4 years ago