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natita [175]
2 years ago
8

A typical reaction time to get your foot on the brake in your car is 0.2 second. If you are traveling at a speed of 60 mph (88 f

t/s or 26.8 m/s), what distance will your car travel during this reaction time? Express your answers in miles, feet and meters.
Physics
2 answers:
Nataly [62]2 years ago
6 0

Answer:

distance in meters= 5.36

distance in feet= 17.6

distance in miles= 3.33*10⁻³

Explanation:

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

Then:

x=\frac{c*b}{a}

Then the rule of three can be applied in the following way, taking into account the speed of 26.8 \frac{m}{s}: if in 1 second you travel 26.8 meters, in the reaction time 0.2 seconds what is the distance traveled?

distance in meters=\frac{0.2 seconds*26.8 meters}{1 second}

<u><em>distance in meters= 5.36</em></u>

Next, the rule of three can be applied in the following way, taking into account the speed of 88 \frac{ft}{s}: if in 1 second you travel 88 feet, in the reaction time 0.2 seconds what is the distance traveled?

distance in feet=\frac{0.2seconds*88 feet}{1 second}

<u><em>distance in feet= 17.6</em></u>

Finally, taking into account that 1 hour is 3600 seconds, the rule of three can be applied in the following way, taking into account the speed of 60 mph: if in 3600 seconds you travel 60 miles, in the reaction time 0.2 seconds what is the distance traveled?

distance in miles=\frac{0.2 seconds*60miles}{3600 seconds}

<u><em>distance in miles= 3.33*10⁻³</em></u>

katrin [286]2 years ago
5 0

Explanation:

Not enough information. It really depends on the technical details of the car ( the data provided is offering just the human factor of the reaction, not the time for getting the impulse through when using the breaks

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A golfer hits a golf ball at an angle of 25.0° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is
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<u>Answer:</u>

 Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

         We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

         Considering upward vertical motion of projectile.

         In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.

        0 = u sin θ - gt

         t = u sin θ/g

    Total time for vertical motion is two times time taken for upward vertical motion of projectile.

    So total travel time of projectile = 2u sin θ/g

Horizontal motion:

  We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g

 So range of projectile,  R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}

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     We have equation of motion, v^2=u^2+2as, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

   Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

   0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}

In the give problem we have R = 301.5 m,  θ = 25° we need to find H.

So  \frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g

Now we have H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m

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2 years ago
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