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omeli [17]
3 years ago
11

Lori’s family is on a road trip. They split their drive into the five legs listed in the table. Find the average velocity for ea

ch leg of the trip. Then arrange the legs of the trip from lowest velocity to highest.

Physics
2 answers:
NARA [144]3 years ago
4 0
It's not possible to answer the question exactly the way it's written.
That's because we don't know anything about the direction they
drive at any time during the trip. 

You see, "velocity" is not just a word that you use for 'speed' when
you want to sound smart and technical, like this question is doing. 
"Velocity" is a quantity that's made up of speed AND THE DIRECTION
of the motion.  If you don't know the direction of the motion, then you
CAN'T tell the velocity, only the speed.

Here are the average speeds that Lori's family drove on each leg
of their trip:

Speed = (distance covered) / (time to cover the distance) .

Leg-A:  
Speed = 15km/10min = 1.5 km/min

Leg-B:  
Speed = 20km/15min = (1 and 1/3) km/min

Leg-C  
Speed = 24km/12min = 2 km/min

Leg-D:  
Speed = 36km/9min = 4 km/min

Leg-E: 
Speed = 14km/14min = 1 km/min

From lowest speed to highest speed, they line up like this:

[Leg-E] ==> [Leg-B] ==> [Leg-A] ==> [Leg-C] ==> [Leg-D]
  1.0 . . . . . . . . 1.3 . . . . . . . 1.5 . . . . . . . 2.0 . . . . . . . 4.0 . . . . km/minute   

Whoever drove Leg-D should have been roundly chastised
and then abandoned by the rest of the family.  36 km in 9 minutes
(4 km per minute) is just about 149 miles per hour !  
OleMash [197]3 years ago
3 0

Answer:

E,B,A,C,D

Explanation:

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A rock is thrown off a 50.0 m high cliff. How fast must the rock leave the cliff top to land on level ground below, 90 m from th
blagie [28]

Answer:

The rock must leave the cliff at a velocity of 28.2 m/s

Explanation:

The position vector of the rock at a time t can be calculated using the following equation:

r = (x0 + v0x · t, y0 + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.

When the rock reaches the ground, the position vector will be (see r1 in the figure):

r1 = (90 m, -50 m)

Then, using the equation of the vector position written above:

90 m = x0 + v0x · t

-50 m = y0 + 1/2 · g · t²

Since x0 and y0 = 0:

90 m = v0x · t

-50 m = 1/2 · g · t²

Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:

-50 m = 1/2 · g · t²

-50 m = -1/2 · 9.81 m/s² · t²

-50 m / -1/2 · 9.81 m/s² = t²

t = 3.19 s

Now, using the equation of the x-component of r1:

90 m = v0x · t

90 m = v0x · 3.19 s

v0x = 90 m / 3.19 s

v0x = 28.2 m/s

8 0
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