(College physics) How do I calculate the agerage speed of 2 trips ( 40km/h and 60km/h (ans not 50km/h?

Question

(College physics) How do I calculate the agerage speed of 2 trips ( 40km/h and 60km/h (ans not 50km/h?

Answer 1

Speed is scalar. Scalars are quantities with only magnitude. The direction does not matter. If you are on the highway whether traveling 100 km/h south or 100 km/h north, your speed is still 100 km/h. Other examples of scalar quantities are shoe size, mass, area, energy.

(average speed) = (total distance)÷(total time)

Example 1: If someone walked 400 m in a straight line in 5 min, their average speed would be (400 m)÷(5 min) = 80 m/min. If the same person walked 100 m [North] then 300 m [South] in 5 minutes, their average speed would still be (400÷5) = 80 m/min. If that person walked 100 m [E] in .75 min, 100 m [N] in 1.50 min, 100 m [W] in 1.00 min and finally 100 m [S] in 1.75 min, their average speed would be (400 m)÷(5 min) = 80 m/min.

Example 2: You drive a car for 2.0 h at 40 km/h, then for another 2.0 h at 60 km/h. a.What is your average speed? b.Do you get the same answer if you drive 100 km at each of the two speeds?

Answer:

a. The total distance driven = [(2 h )(40 km /h) + (2 h)( 60 km/h)] = 200 km

The total time = 2 + 2 = 4 h

average speed = (200 km)/(4 h) = 50km/h

b. total distance = 100 + 100 = 200 km

total time = [(100 km)/(40 km/h) + (100 km)/ 60 km/h)] = 4.17 h

average speed = (200 km)/(4.17 h) = 48 km/h

(average velocity) = displacement ÷ time.

Velocity is a vector. Both direction and quantity must be stated. It one train has a velocity of 100km/h north, and a second train has a velocity of 100km/h south, the two trains have different velocities, even though their speed is the same. Other examples of vectors are force, and field intensity.

Example 3: if a person walked 400 m in a straight line in 5 min, that person's velocity would be (400 m [forward])÷(5 min) = 80 m/min [forward] .

If the same person walked 100 m [North] then 300 m [South] in 5 minutes, we first find their displacement.

displacement = 200 m [S]

velocity = 200÷5 = 40 m/min [S]

If that person walked 100 m [E] in .75 min, 100 m [N] in 1.50 min, 100 m [W] in 1.00 min and finally 100 m [S] in 1.75 min, that person would end up back where they started. Since their displacement is zero, Their velocity is zero.

Remember,

(average velocity) = displacement ÷ time.

Example 4: A hiker traveled 80.0 m [S] at 1.00 m/s, then 80.0 m [S] at 5.00 m/s. What is the hiker's average velocity?

Answer:

displacement = 160.0 m [S]

time for the first part is 80.0÷1.00= 80.0 s, time for the second part is 80.0 m ÷ 5.00 m/s = 16.0 s.

Total time = 80.0+16.0 = 96.0 s

Therefore, the velocity is (160.0 m [S])÷96.0 s = 1.67 m/s [S]

Example 5: A train on a straight track traveled 60.0 km/h [E] for 2.00 h, stopped for 15 min, then traveled 100.0 km [W] at 133 km/h.

a. What was the train's average speed for the whole trip?

a. What was the train's average velocity for the whole trip?

Answer:

a. To find average speed, we need total distance and total time.

During the first part of the trip, the train covered 60.0x2.00 = 120 km in 2.00 h.

During the second part of the trip the train traveled 0.00 km in 0.25 h.

During the third part of the trip, the train traveled 100.0 km in 0.75 h.

In total the train traveled 220 km in 3 .00 h.

Average speed = (220 km)÷3.00 = 73.3 km/h

b. To find average velocity, we need displacement and total time.

During the first part of the trip, the train covered 60.0x2.00 = 120 km [E] in 2.00 h.

During the second part of the trip the train traveled 0.00 km in 0.25 h.

During the third part of the trip, the train traveled 100.0 km [W] in (100.0 km ÷ 133 km/h) = 0.75 h.

The train's displacement was (120-100) = 20 km [E] in 3 .00 h.

Average velocity = (20 km [E])÷3.00h = 6.7 km/h [E]

Example 6: A runner covers one lap of a circular track 40.0 m in diameter in 62.5 s. For that lap, what were her average speed and average velocity?

answer:

average speed = (total distance)/(total time) = (π*40.0)/(62.5) = 2.01m/s

average velocity = displacement/time = 0/62.5 = 0 m/s

acceleration = (change in velocity) ÷ time.

Acceleration is a vector when it refers to the rate of change of velocity. Acceleration is scalar when it refers to rate of change of speed. A car slowing down to stop at a stop sign is accelerating because its speed is changing. We might refer to this type of acceleration as deceleration or negative acceleration. A car going at a constant speed around a curve is still accelerating because its direction is changing.

Example 7: A pitcher delivers a fast ball with a velocity of 43 m/s to the south. The batter hits the ball and gives it a velocity of 51m/s to the north. What was the average acceleration of the ball during the 1.0ms when it was in contact with the bat?

answer:

acceleration = (vf - vi)/t = ( 51m/s to the north - 43 m/s to the south)/(1.0x10-3s)

Letting south be positive and north negative yields

acceleration = ( -51m/s - 43 m/s )/(1.0x10-3s) = -94000 m/s/s

acceleration = 94000 m/s/s to the north

Answer 2

maybe u need to add the rotational speed of the earth, or give final distance from origin.

Answer 3

without time and distance known there is nothing. youalready have the answer.

Answer 4

Give us more details, otherwise the answer is 50km/h!!