Question

4.) In coming to a stop, a car leaves skid marks 80m long on the highway. Assuming a DEceleration of 7.00 m/s^2 (m/s/s), estimate the speed of the car just before breaking.

6.) A car traveling @ 90km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80m. What was the average acceleration of the driver during the collision? Express answer in terms of ("g's") where 1.00 g = 9.80 m/s^2 OR (m/s/s)

Answer 1

I found an example like #6...

A car traveling at 95m/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80m. What was the average acceleration of the driver during the collision? Express the answer in terms of "g's," where 1.00g = 9.80m/s2.

_________________________________________________
Here's the WORK:

v2 = vo2 + 2aav(x2 - x1);

0 = [(95km/h)(3.6ks/h)]2 + 2aav(0.80m), which gives aav = -4.4 x 102m/s2.

The number of g's is
| aav | = (4.4 x 102m/s2)/[(9.80m/s2)/g] = 44g.

**I DON'T GET WHERE THE (3.6 ks/h) CAME FROM. HOW DID THE PERSON GET 3.6?? **

Answer 2

4) x=(1/2)a*t^2
t=sqrt(2x/a)
plug t into --> v=a*t

6) 90km/h = 25m/s
v=at
a=v/t --> a=25/t
plug in a and solve --> x=x+vt+(1/2)at^2
0.8 = 25t - (1/2)(25/t)t^2
That gives you t which you can then plug back into a=v/t to get acceleration. You divide that by 9.8m/s^2 to give you g's.

Answer 3

LOST! I'm confused putting the number in!!
Test tmrw!!!

Answer 4

Ok, starting from 0.8=25t-(1/2)(25/t)t^2

0.8 = 25t - (25t^2/2t)

0.8 = 25t - 12.5t

0.8 = 12.5t

t = 0.8/12.5

Plug that back into our first equation:

velocity = acceleration * time

a = v / t

a = 25m/s / (0.8/12.5)

a / 9.8 = g force

Plug in values and and prosper.



You should get a g force of approx 39.86

Answer 5

Shouldn't my answer be negative 39.86?

Answer 6

A car is traveling at a speed of 80km/h on an icy highway. The driver swerves while applying the brakes to avoid a collision. if the Yaw mark that the car leaves on the icy asphalt has a radius of 270 m, determine the drag factor of the icy highway. Round to the nearest hundredth.

Answer 7

s = sqrt127fr

Where,
s = the speed (km/h) the vehicle was travelling when the brakes were applied
f = the drag factor of the surface
r = radius (m) of the curved Yaw mark

A car is traveling at a speed of 80km/h on an icy highway. The driver swerves while applying the brakes to avoid a collision. if the Yaw mark that the car leaves on the icy asphalt has a radius of 270 m, determine the drag factor of the icy highway. Round to the nearest hundredth.

Answer 8

sqrt8

Answer 9

entire radical next to its equivalent mixed radical is 2sqrt3

Answer 10

9 square root of 14 and square root of 14 are like terms. select one true or false

Answer 11

When you add 6 square root of 3 and 5 square root of 3, you end up with __√ ___

Answer 12

When you simplify 6 square root of 3 minus 3 square root of 9 plus 2 square root of 27, the result is ___ √___ - ___.

Answer 13

Simplify (3 square root of 32) (4 square root of 24).

Answer 14

Simplify (8 square root of 6) (2 square root of 2 minus square root of 3)

Answer 15

Simplify (square root of 6 minus 5) (square root of 10 minus 5)

Answer 16

sqrt 150 / sqrt 6

Answer 17

\frac{4\sqrt{16}}{6\sqrt{3}}

Answer 18

Rationalize the denominator sqrt 6 minus 4 fraction sqrt 5

Answer 19

Rationalize the denominator sqrt 6 minus 4/sqrt 5

Answer 20

Simplify 9 square root of a plus square root of 20 a end root minus square root of 5 a end root plus 7 square root of 9 a end root

Answer 21

To solve for the drag factor of the icy highway, we can use the equation:

drag factor = (velocity^2) / (radius * acceleration)

First, let's convert the velocity from km/h to m/s:

80 km/h = 80 * (1000 m / 3600 s) = 22.22 m/s

Next, let's plug in the values into the equation:

drag factor = (22.22^2) / (270 * acceleration)

To determine the acceleration, we need to make use of the centripetal acceleration formula:

acceleration = (velocity^2) / radius

acceleration = (22.22^2) / 270

Now we can substitute the acceleration value back into the drag factor equation:

drag factor = (22.22^2) / (270 * [(22.22^2) / 270])

Simplifying:

drag factor = 1

Therefore, the drag factor of the icy highway is 1.00.

Answer 22

To find the drag factor of the icy highway, we can use the equation:

s = √(127f * r)

Given that the speed (s) is 80 km/h and the radius (r) of the yaw mark is 270 m, we can rearrange the equation to solve for the drag factor (f):

f = (s^2) / (127 * r)

First, let's convert the speed from km/h to m/s:

s = 80 * (1000 m / 3600 s)
s = 22.22 m/s

Now we can plug in the values into the equation:

f = (22.22^2) / (127 * 270)
f ≈ 0.045

Therefore, the drag factor of the icy highway is approximately 0.045.

Answer 23

The square root of 8 is approximately 2.828.

Answer 24

To rewrite √8 as a mixed radical, we can break it down into its simplest form by factoring out perfect square factors from the radicand.

√8 = √(4 * 2)

Since 4 is a perfect square, we can take its square root out of the radical:

√(4 * 2) = 2√2

Therefore, the expression 2√3 is not equal to √8.

Answer 25

False.

9√14 and √14 are not like terms because they have different coefficients. Like terms have the same variable(s) and exponent(s), but the coefficients can be different. In this case, the coefficient of 9 in 9√14 makes it different from √14.

Answer 26

When you add 6 square root of 3 and 5 square root of 3, you end up with:

(6√3) + (5√3) = 11√3

Answer 27

To simplify the expression 6√3 - 3√9 + 2√27, we can first simplify the square roots of 9 and 27:

√9 = 3
√27 = 3√3

Now we substitute these values back into the expression:

6√3 - 3√9 + 2√27 = 6√3 - 3(3) + 2(3√3)
= 6√3 - 9 + 6√3
= (6√3 + 6√3) - 9
= 12√3 - 9

The simplified expression is:
12√3 - 9.

Answer 28

To simplify (3√32)(4√24), we can first simplify the square roots of 32 and 24:

√32 = 4√2
√24 = 2√6

Now we substitute these simplified square roots back into the expression:

(3√32)(4√24) = 3(4√2)(4)(2√6)
= 12√2 * 4 * 2√6
= 12 * 4 * √2 * 2√6
= 96√2√6

Since √2 and √6 cannot be further simplified, we multiply the coefficients:

96 * 2 * √2 * √6
= 192√12

To simplify further, we can find the square root of 12:

√12 = √(4 * 3) = 2√3

Now we substitute back into the expression:

192√12 = 192(2√3)
= 384√3

Therefore, the simplified expression is 384√3.

Answer 29

To simplify (8√6)(2√2 - √3), we can distribute the 8√6 to both terms inside the parentheses:

(8√6)(2√2 - √3) = 16√12 - 8√18

Next, we can simplify the square roots of 12 and 18:

√12 = 2√3
√18 = 3√2

Substituting these simplified square roots back into the expression:

16√12 - 8√18 = 16(2√3) - 8(3√2)
= 32√3 - 24√2

Therefore, the simplified expression is 32√3 - 24√2.

Answer 30

To simplify (√6 - 5)(√10 - 5), we can use the distributive property:

(√6 - 5)(√10 - 5) = √6 * √10 - 5√10 - 5√6 + 25

Now, let's simplify each term individually:

√6 * √10 = √(6 * 10) = √60 = 2√15

-5√10 and -5√6 cannot be simplified further.

Finally, combining all the simplified terms:

(√6 - 5)(√10 - 5) = 2√15 - 5√10 - 5√6 + 25

Therefore, the simplified expression is 2√15 - 5√10 - 5√6 + 25.

Answer 31

To simplify √150 / √6, we can divide the square roots by rationalizing the denominator:

√150 = √(25 * 6) = 5√6

Now, we can rewrite the expression:

(5√6) / √6

Since the square root of 6 appears in both the numerator and denominator, they will cancel out:

(5√6) / √6 = 5

Therefore, the simplified expression is 5.

Answer 32

To simplify the expression $\frac{4\sqrt{16}}{6\sqrt{3}}$, we can simplify the square roots and simplify the fraction:

$\frac{4\sqrt{16}}{6\sqrt{3}} = \frac{4 \cdot 4}{6\sqrt{3}} = \frac{16}{6\sqrt{3}}$

Next, we can simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2:

$\frac{16}{6\sqrt{3}} = \frac{8}{3\sqrt{3}}$

Finally, to rationalize the denominator, we can multiply the numerator and denominator by $\sqrt{3}$:

$\frac{8}{3\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{8\sqrt{3}}{3 \cdot 3} = \frac{8\sqrt{3}}{9}$

Therefore, the simplified expression is $\frac{8\sqrt{3}}{9}$.

Answer 33

To rationalize the denominator $\sqrt{6} - \frac{4}{\sqrt{5}}$, we can multiply the numerator and denominator by the conjugate of the denominator, which in this case is $\sqrt{6} + \frac{4}{\sqrt{5}}$:

$\left(\sqrt{6} - \frac{4}{\sqrt{5}}\right) \cdot \left(\frac{\sqrt{6} + \frac{4}{\sqrt{5}}}{\sqrt{6} + \frac{4}{\sqrt{5}}}\right)$

Using the distributive property on the numerator:

$\left(\sqrt{6} - \frac{4}{\sqrt{5}}\right) \cdot \left(\frac{\sqrt{6}}{\sqrt{6}} + \frac{4}{\sqrt{5}} \cdot \frac{5}{\sqrt{5}}\right)$

Simplifying the numerator:

$\sqrt{6} \cdot \sqrt{6} + \frac{4}{\sqrt{5}} \cdot \frac{5}{\sqrt{5}} \cdot \sqrt{6} - \frac{4}{\sqrt{5}} \cdot \frac{4}{\sqrt{5}}$

$6 + \frac{4}{\sqrt{5}} \cdot \sqrt{6} - \frac{4}{5}$

$6 + \frac{4 \sqrt{6}}{\sqrt{5}} - \frac{4}{5}$

Now, the denominator becomes a rational number:

$\sqrt{6} + \frac{4 \sqrt{6}}{\sqrt{5}} - \frac{4}{5}$

Combining like terms:

$\sqrt{6} \left(1 + \frac{4}{\sqrt{5}}\right) - \frac{4}{5}$

$\sqrt{6} \left(\frac{\sqrt{5} + 4}{\sqrt{5}}\right) - \frac{4}{5}$

Simplifying further:

$\frac{\sqrt{6}(\sqrt{5} + 4)}{\sqrt{5}} - \frac{4}{5}$

Now, we rationalized the denominator of the original expression and simplified as much as possible.

Answer 34

To rationalize the denominator $\frac{\sqrt{6} - \frac{4}{\sqrt{5}}}{\sqrt{5}}$, we can multiply the whole fraction by $\frac{\sqrt{5}}{\sqrt{5}}$:

$\frac{\sqrt{6} - \frac{4}{\sqrt{5}}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$

Simplifying the numerator, we get:

$\frac{\sqrt{6} \cdot \sqrt{5} - \frac{4}{\sqrt{5}} \cdot \sqrt{5}}{\sqrt{5}}$

$\frac{\sqrt{30} - \frac{4 \cdot \sqrt{5}}{\sqrt{5}}}{\sqrt{5}}$

$\frac{\sqrt{30} - 4}{\sqrt{5}}$

Therefore, the rationalized form of the fraction is $\frac{\sqrt{30} - 4}{\sqrt{5}}$.

Answer 35

To simplify the expression $\sqrt{9a} + \sqrt{20a} - \sqrt{5a} + 7\sqrt{9a}$, we can first simplify each square root term individually:

$\sqrt{9a} = 3\sqrt{a}$

$\sqrt{20a} = 2\sqrt{5a}$

$\sqrt{5a}$ cannot be simplified further.

$7\sqrt{9a} = 7(3\sqrt{a}) = 21\sqrt{a}$

Now, we can substitute these simplified terms back into the original expression:

$3\sqrt{a} + 2\sqrt{5a} - \sqrt{5a} + 21\sqrt{a}$

Combining like terms:

$(3\sqrt{a} + 21\sqrt{a}) + (2\sqrt{5a} - \sqrt{5a})$

$24\sqrt{a}$

Therefore, the simplified form of the expression is $24\sqrt{a}$.