You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has stopped due to an accident up ahead. You step on your brakes to slow down as quickly as possible. Assume that you to slow down to 30 mph in about 5 seconds. A) With this same average acceleration, how much longer would it take you to stop?B) What total distance would you travel from when you first apply the brakes until the car stops?

Answers

Answer 1
Answer:

A.The time taken for the car to stop is 8.75 s

B.The distance travelled when the brakes were applied till the car stops is 136.89 m

A. Determination of the time taken for the car to stop.

  • We'llbegin bycalculatingthedecelerationof thecar

Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s

Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s

Time (t) = 5 s

Deceleration (a) =?

a \:  =  (v \:  - u)/(t)  \n  \n a =  (13.41 - 31.29)/(5)  \n  \n a \:  =  ( - 17.88)/(5)  \n  \n

a = –3.576 m/s²

  • Finally,we shall determine the time taken for the car to stop.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Time (t) =?

v \:  = u \:  + at \n 0 \:  = 31.29 \:  +  \: ( - 3.576 * t) \n 0 \:  = 31.29 \:  - 3.576 * t \n collet \: like \: terms \n 0 - 31.29 \:  = - 3.576 * t  \n - 31.29 \:  = - 3.576 * t  \n divide \: both \: side \: by \:  - 3.576 \n t \:  =  (- 31.29)/(- 3.576)  \n

t = 8.75 s

Thus, the time taken for the car to stop is 8.75 s

B.Determination of the total distance travelled when the brakes were applied.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Distance (s) =?

{v}^(2)  =  {u}^(2)  + 2as \n {0}^(2)  =  {31.29}^(2)  + (2 *  - 3.576 * s) \n 0  = 979.0641   - 7.152 s \n collect \: like \: terms \n 0  -  979.0641  =  - 7.152 s \n -  979.0641  =  - 7.152 s \n divide \: both \: side \: by \: - 7.152 \n s  =  (-  979.0641)/(- 7.152)  \n  \n

s = 136.89 m

Therefore, the total distance travelled by the car when the brakes were applied is 136.89 m

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Answer 2
Answer:

Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity=70 mph\approx 31.29 m/s

velocity after 5 s is 30 mph\approx 13.41 m/s

Therefore acceleration during these 5 s

a=(v-u)/(t)

a=(13.41-31.29)/(5)=-3.576 m/s^2

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

0=31.29-3.576* t

t=(31.29)/(3.576)=8.75 s

(b)total distance traveled before stoppage

v^2-u^2=2as

0^2-31.29^2=2* (-3.576)\cdot s

s=136.89 m


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Answers

Answer:

Explanation:

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Answers

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Answers

Answer:

F_(AB) = 172.1356\nF_(AC) = 258.2033\nF_(AD) = 368.8004

Explanation:

Using the diagram (see attachment) we extract the following position vectors:

Vector (OA) = 6i + 0j + 0k \nVector (OB) = 0i + 3j + 2k \nVector (OC) = 0i - 2j + 3k

Next step is to find unit vectors u_(AB) ,u_(AC), u_(AD), u_(AE) as follows:

u_(AB) = (vector(AB))/(magnitude(AB)) \n= \frac{OB - OA}{magnitude({vector(OB - OA))} }\n=(-6i +3j+2k)/(√(6^2 + 3^2+2^2) ) \n\n=-0.857 i +0.429j+0.286k\n\nu_(AC) = (vector(AC))/(magnitude(AC)) \n= \frac{OC - OA}{magnitude({vector(OC - OA))} }\n=(-6i -2j-3k)/(√(6^2 + 2^2+3^2) ) \n\n=-0.857 i -0.286j+0.429k\n\nu_(AD) = +1i\nu_(AC) = -1k

Using the diagram we find the corresponding vectors Forces:

F_(AB) = F_(AB) i + F_(AB)j +F_(AB)k\nF_(AC) = F_(AC) i + F_(AC)j +F_(AC)k\nF_(AD) = F_(AD) i + F_(AD)j +F_(AD)k\nW = -160 k

Equation of Equilibrium:

Sum of forces = 0\nF_(AB). u_(AB) + F_(AC).u_(AC) + F_(AD).u_(AD) + W = 0\n(-0.857F_(AB)i + 0.429F_(AB)j +0.286F_(AB)k) + (-0.857F_(AC)i - 0.286F_(AC)j +0.429F_(AC)k) + (+1F_(AD) i)  + (-160k) = 0

Comparing i , j and k components as follows:

-0.857F_(AB) -0.857F_(AC)  +1F_(AD)  = 0\n+ 0.429F_(AB) - 0.286F_(AC) = 0\n+0.286F_(AB) +0.429F_(AC)  =  160

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F_(AB) = 172.1356\nF_(AC) = 258.2033\nF_(AD) = 368.8004

Which of the following is correct? *PLEASE HELP MEEEE
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Answers

Answer:

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Explanation:

If you do not trust me look it up

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Answers

We are given:

The tuning fork vibrates at 15660 oscillations per minute

Period of one back-and forth movement:

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Answers

Answer:

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Solution:

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Now,

d = (\mu_(o)I)/(2\pi B)

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