Ask question

Ask question

All categories

julia-pushkina [17]

3 years ago

8

You are driving downhill on a rural road with a 3% grade at a speed of 45 mph. While playing on the side of the road, a child ac

cidentally runs onto the roadway to pick up a ball. At what distance you must first see the child to stop your vehicle before reaching the child? How would this distance be different if you were driving on the same road and at the same speed, but going uphill? Use t

1 answer:

Dennis_Churaev [7]3 years ago

5 0

Answer: a) 95.07m b) 81.88 m

Explanation:

a)

For finding the distance when vehicle is going downhill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1 + $\frac{(0.28*72.4)^{2} }{2*9.81*(0.31-0.03)}$

Stop sight distance= 95.07 m

b)

For finding the distance when vehicle is going uphill we have the formula as:

Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)

Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31

Reaction time= 0.28

So putting values we get

Stop sight distance= 0.28*72.4 *1 + $\frac{(0.28*72.4)^{2} }{2*9.81*(0.31+0.03)}$

Stop sight distance= 81.88 m

You might be interested in

the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity

NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

$\int\limits^s_0 {a} \, ds = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\$

Remember that

$v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2$

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0

3 years ago

Help me please i dont understand

solniwko [45]

Answer:

It is b

Explanation:

The question tells you that the distance was increased 3 times the original distance, which means it was moved 3 times

5 0

3 years ago

1. Which mathematical representation correctly identifies impulse?

horsena [70]

Answer:

1. B. Impulse = Force × Time

2. A. The momentum of each ball changes, and the total momentum stays the same

3. -55 kg·m/s

4. B. 3.5 kg

5. C. 6.3 m/s

Explanation:

1. The impulse is the momentum change of an object due to a force applied for a given period

2. Given that the objects collide, and the force of the 3 kg mass moving with 24 kg·m/s acts on the 1 kg mass while the total momentum is conserved;

The stationary ball of mass 1 kg begins to moves at certain velocity after collision and therefore changes momentum, while the velocity of the ball of mass 3.0 kg reduces and the total combined momentum of the two balls in the closed system remains the same

3. By the principle of conservation of linear momentum, we have;

The sum of the momentum before the collision = The sum of the momentum after collision

Given that the objects move together after the collision, the total momentum is therefore;

Total momentum = 110 kg·m/s + -65 kg·m/s + -100 kg·m/s = 110 kg·m/s - 65 kg·m/s - 100 kg·m/s = -55kg·m/s

4. Given that the final velocity of the two objects (m₁ + m₂) combined = 50 m/s

Where;

m₁ = The mass of the first object

m₂ = The mass of the second object

The total momentum of the system = 250 kg·m/s

From momentum = Mass × Velocity, we have;

Mass = Momentum/Velocity = 250 kg·m/s/(50 m/s) = 5.0 kg

The mass (m₁ + m₂) = 5.0 kg

Given that m₁ = 1.5 kg, we have;

m₂ = 5.0 kg - m₁ = 5.0 kg - 1.5 kg = 3.5 kg

The mass of the second object = 3.5 kg

5. The mass of the cue stick = 0.5 kg

The velocity of the cue stick = 2.5 m/s

The mass of the ball = 0.2 kg

The initial velocity of the ball = 0 m/s

Given that total initial momentum = Total final momentum, we have;

0.5 kg × 2.5 m/s + 0.2 kg × 0 = 0.2 kg × v + 0.5 kg × 0

0.5 kg × 2.5 m/s = 0.2 kg × v

v = (0.5 kg × 2.5 m/s)/(0.2 kg) = 6.25 m/s ≈ 6.3 m/s

3 0

3 years ago

What two things are necessary for work to be done on an object?

NemiM [27]

Answer: Force and Movement

Explanation:

The first is that the object moves. The second is that a force must act on the object in the direction the object moves.

6 0

3 years ago

two ice skaters push each other, and move in opposite directions. the mass of one is 75 kg, and his speed is 2 m/s. if the mass

sattari [20]

momentum conservation

75x2 = 30 x her speed

75x2/30=150/30=5m/s

7 0

3 years ago