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Alexeev081 [22]
2 years ago
8

to plan ahead using reps and checks, choose the minimum space cushion needed to maintain following distance traveling at 25 mph

(for a cargo van)?
Physics
1 answer:
Sveta_85 [38]2 years ago
6 0

The kinematics allows finding the answer for the shortest distance that must exist between the two vehicles so that the collision does not occur during the reaction time is 13.75 ft.

Kinematics studies the movement of bodies, establishes relationships between the position, speed and acceleration of bodies.

the reaction time. The human being needs some time to process the information that comes to him and you take some action, this time is defined as the reaction time between seeing an image and taking some action in response,  in the case of a driver in good condition this reaction time is 0.75 s, in the case of people with alcohol in the blood it can go up to 2.0 s.

The safe distance is the distance between the two vehicles so that when the front vehicle applies the brakes, during the reaction time the rear vehicle does not collide with the front vehicle.

Let's start by reducing the magnitudes

          v₀ = 25 m / h (\frac{5280 ft}{1 mi}) ( \frac{1 h}{3600s})

          v₀ = 36.66 ft / s

We look for the braking acceleration for the reaction time of a normal driver (t = 0.75 s)

           v = v₀ - a t

final velocity is zero

           0 = v₀ - a t

           a = \frac{v_o^2}{t}

           a = \frac{36.666}{0.75}

           a = 48.888 ft / s²

we look for the distance traveled, for the final velocity zero

           v = v_o^2 - 2 a x

          0 = v_o^2 - 2ax

           x = \frac{v_o^2}{2a}  

           x = \frac{36.66^2}{2 \ 48.88}

           x = 13.75 ft

reaction time

Usings kinematic we can find the shortest distance that must exist between the two vehicles so that the collision does not occur during the  is 13.75 ft.

Learn more here:  brainly.com/question/24578621

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amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

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\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

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A)6.15 cm to the left of the lens

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<u>Given the following data;</u>

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