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3 years ago

6

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 38 m in front of you.Your reaction time

before stepping on the brakes is 0.50 s ,and the maximum deceleration of your car is 10 m/s^2 what is the maximum speed you could have and still not hit the deer

1 answer:

shtirl [24]3 years ago

3 0

Consider the motion of the car before brakes are applied:

v₀ = maximum initial velocity of the car before the brakes are applied

t = reaction time = 0.50 s

x₀ = distance traveled by the car before brakes are applied

since car moves at constant speed before brakes are applied

Using the equation

x₀ = v₀ t

x₀ = v₀ (0.50)

Consider the motion after brakes are applied :

v₀ = initial velocity of the car before the brakes are applied

a = acceleration = - 10 m/s²

v = final velocity of the car after it comes to stop = 0 m/s

x = stopping distance = initial distance - distance traveled before applying the brakes = 38 - x₀ = 38 - v₀ (0.50)

Using the equation

v² = v²₀ + 2 a x

inserting the values

0² = v²₀ + 2 (- 10) (38 - v₀ (0.50))

v²₀ = 20 (38 - v₀ (0.50))

v₀ = 23 m/s

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3 years ago

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

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Answer:

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(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

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The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

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(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

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(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

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3 years ago

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2 years ago

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To calculate the time, we use kinematic expression for constant speed:

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