Ask question

Ask question

All categories

3 years ago

11

Tests reveal that a normal driver takes about 0.75 s before he orshecan react to a situation to avoid a collision. It takes abou

t 3 sfora driver having 0.1% alcohol in his system to do the same. Ifsuchdrivers are traveling on a straight road at 44 ft/s and theircars can decelerate at 2 ft/s^2, determine the shortest stoppingdistanced for each from the moment they see the pedestrians.

1 answer:

stellarik [79]3 years ago

6 0

Answer:

The shortest stopping distance for the normal driver is 517 ft

The shortest stopping distance for the alcoholic driver is 616 ft.

Explanation:

The data, we have is:

Vi = Initial Speed = 44 ft/s

Vf = Final Speed = 0 ft/s (Since, the car finally stops)

a = deceleration = - 2 ft/s²

t = time taken to stop after applying brakes = ?

First, we calculate the time taken by the car to stop after applying the brakes:

Using 1st equation of motion:

Vf = Vi + at

t = (Vf - Vi)/a

t = (0 ft/s - 44 ft/s)/(-2 ft/s²)

t = 22 sec

Now, the stopping distance after applying brakes (S1) is given by 2nd equation of motion:

S1 = Vi t + (1/2)at²

S1 = (44 ft/s)(22 s) + (1/2)(-2 ft/s²)(22 s)²

S1 = 484 ft

<u>FOR NORMAL DRIVER</u>:

Since, the driver takes 0.75 s to respond to a situation, so the distance traveled in this time (S2) is given by:

S2 = Vt

S2 = (44 ft/s)(0.75 s)

S2 = 33 ft

Thus, the total stopping distance for a normal driver is:

S = S1 + S2

S = 484 ft + 33 ft

<u>S = 517 ft</u>

<u>FOR ALCOHOLIC DRIVER</u>:

Since, the driver takes 0.75 s to respond to a situation, so the distance traveled in this time (S2) is given by:

S2 = Vt

S2 = (44 ft/s)(3 s)

S2 = 132 ft

Thus, the total stopping distance for a normal driver is:

S = S1 + S2

S = 484 ft + 132 ft

<u>S = 616 ft</u>

You might be interested in

The 150-lb man sits in the center of the boat, which has a uniform width and a weight per linear foot of 3 lb>ft. Determine t

irina1246 [14]

Answer:

M = 281.25 lb*ft

Explanation:

Given

W<em>man</em> = 150 lb

Weight per linear foot of the boat: q = 3 lb/ft

L = 15.00 m

M<em>max</em> = ?

Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):

∑ Fy = 0 (+↑) ⇒ q'*L - W - q*L = 0

⇒ q' = (W + q*L) / L

⇒ q' = (150 lb + 3 lb/ft*15 ft) / 15 ft

⇒ q' = 13 lb/ft (+↑)

The free body diagram of the boat is shown in the pic.

Then, we apply the following equation

q(x) = (13 - 3) = 10 (+↑)

V(x) = ∫q(x) dx = ∫10 dx = 10x (0 ≤ x ≤ 7.5)

M(x) = ∫10x dx = 5x² (0 ≤ x ≤ 7.5)

The maximum internal bending moment occurs when x = 7.5 ft

then

M(7.5) = 5(7.5)² = 281.25 lb*ft

8 0

3 years ago

1. A 260 ft (79.25 m) length of size 4 AWG uncoated copper wire operating at a tem-

Murljashka [212]

A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.

Explanation:

From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).

To find the resistance of 260 ft (79.25 m) of size 4 AWG,

R= K * L/ A

K = 0.0214 ohm mm²/m

L = 79.25 m

A = 21.2 mm²

R = 0.0214 * $\frac{79.25}{21.2}$

= 0.0214 * 3.738

= 0.0792 ohm.

Thus the resistance of uncoated copper wire is 0.0792 ohm

5 0

3 years ago

You are ordering steel cable for a 250 foot long zip-line you are building in your back yard. The cable can be ordered in diamet

Margarita [4]

Answer:

If i am correct It should be 1/4 of an inch

Explanation:

Sorry but i can't quite explain

3 0

3 years ago

Read 2 more answers

The time factor for a doubly drained clay layer

Margarita [4]

Answer with Explanation:

Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is

$T_v=\frac{\pi }{4}(\frac{U}{100})^2$

Solving for 'U' we get

$\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%$

Since our assumption is correct thus we conclude that degree of consolidation is 50.46%

The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2

i) $\frac{z}{H}=0.25=U=0.71$ = 71% consolidation

ii) $\frac{z}{H}=0.5=U=0.45$ = 45% consolidation

iii) $\frac{z}{H}=0.75U=0.3$ = 30% consolidation

Part b)

The degree of consolidation is given by

$\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm$

Thus a settlement of 50.46 centimeters has occurred

For time factor 0.7, U is given by

$T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59$

thus consolidation of 85.59 % has occured if time factor is 0.7

The degree of consolidation is given by

$\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm$

5 0

3 years ago

A pump with a power of 5 kW (pump power, and not useful pump power) and an efficiency of 72 percent is used to pump water from a

almond37 [142]

Answer:

a) The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump is 245.175 kilopascals.

Explanation:

a) Let suppose that pump works at steady state. The mass flow rate of the water ( $\dot m$ ), in kilograms per second, is determined by following formula:

$\dot m = \frac{\eta \cdot \dot W}{g\cdot H}$ (1)

Where:

$\dot W$ - Pump power, in watts.

$\eta$ - Efficiency, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$H$ - Hydrostatic column, in meters.

If we know that $\eta = 0.72$ , $\dot W = 5000\,W$ , $g = 9.807\,\frac{m}{s^{2}}$ and $H = 25\,m$ , then the mass flow rate of water is:

$\dot m = 14.683\,\frac{kg}{s}$

The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump ( $\Delta P$ ), in pascals, is determined by this equation:

$\Delta P = \rho\cdot g\cdot H$ (2)

Where $\rho$ is the density of water, in kilograms per cubic meter.

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $H = 25\,m$ , then the pressure difference is:

$\Delta P = 245175\,Pa$

The pressure difference across the pump is 245.175 kilopascals.

4 0

3 years ago