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alex41 [277]
3 years ago
9

The two-second rule applies to any speed __________ A. in all weather conditions. B. in ideal weather and road conditions. C. in

the worst weather and road conditions. D. none of the above
Physics
2 answers:
kumpel [21]3 years ago
5 0

Answer:

The answer is A

Explanation:

A large risk of tailgating is the collision avoidance time being much less than the driver reaction time. Driving instructors advocate that drivers always use the "two-second rule" regardless of speed or the type of road. During adverse weather, downhill slopes, or hazardous conditions such as black ice, it is important to maintain an even greater distance.

Flura [38]3 years ago
5 0

Answer:

in ideal weather and road conditions.

Explanation:

Minimum Safe Following Distances

Leave plenty of space between you and the vehicle ahead, including bicycles. If it stops quickly, you will need time to see the danger and stop.

Using the Two-Second Rule

At any speed, you can use the two-second rule to see if you are far enough behind the car in front of you:

Watch the vehicle ahead pass some fixed point - an overpass, sign, fence corner, or other marker.

Count off the seconds it takes you to reach the same spot in the road ("one thousand and one, one thousand and two...").

If you reach the mark before you finish counting, you are following too closely. Slow down and check your following distance again.  

The two-second rule applies to any speed in ideal weather and road conditions. If road or weather conditions are not good, double your following distance. You should also double your following distance when driving a motor home or towing a trailer.

Following Distance For Trucks

A truck or any vehicle towing another vehicle may not follow within 300 feet of another truck or vehicle towing a vehicle. This law does not apply to overtaking and passing, and it does not apply within cities or towns.

You might be interested in
When you apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed ().When you apply a f
elena-s [515]

Answer:

Explanation:

When we apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed. So net force on the block is zero .

It implies that a force ( frictional ) acts on it which is equal to 76 N in opposite direction ( friction )

When we apply  a greater force on it it starts moving with acceleration .

This time kinetic friction acts on it due to rough ground equal to 76 N .This is limiting friction ( maximum friction )

Net force on the body in later case

= 89 - 76

= 13 N

Force by ground on the block in horizontal direction = 76 N ( FRICTIONAL FORCE )

=

3 0
3 years ago
Consider the position vs. time graph below for a woman's movement in a hallway. What is the woman's velocity from 4 to 5 s?
Ksenya-84 [330]

Answer:

The answer is "6\  \frac{m}{s}"

Explanation:

The formula for velocity:

\to \overline{v}={\frac{\Delta x}{\Delta t}}

      =\frac{6}{1}\\\\=6\  \frac{m}{s}

7 0
2 years ago
A motorcycle is speeding along at a velocity of120kph.Then, it changes to a velocity of 150 kph for 2 minutes.What is the motorc
Taya2010 [7]

Answer:

Explanation:

a=v-u/t

a=acceleration

v=final velocity

u=initial velocity

t=tme taken

we need to convert from kph to ms⁻¹

v= 150*1000/60*60= 41.67ms⁻¹

u= 120*1000/60*60= 33.33ms⁻¹

t= 2*60= 120s

a=41.67-33.33/120

a=8.34/120

a=0.0694ms⁻²

8 0
2 years ago
A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a
sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0
3 years ago
A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h
Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft

The minimum stopping distance is 141.78 ft.

4 0
3 years ago
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