Question

At t1, Car A and car B are each located at position xo moving forward at speed v. At t2, car A is located at position 2xo moving forward at speed 3v, while car B is located at position 2xo but is moving backward at speed v. Is the average velocity of car A between t1 and t2greater than, less than, or equal to the average velocity of car B between t1 and t2?

Answer 1

Answer:

the average velocity of car A between t1 and t2greater is greater than the average velocity of B berween t1 and t2

Explanation:

Velocity is displacement over time,

Displacement is the distance covered relative to the initial starting position

For A:

at time ti, A moved from Xo to 2Xo, displacement is 2Xo.

at time t2 a moves with speed 3V, hence, his new position will be 3Xo from 2Xo which will be at 5Xo. A's displacement is 5Xo from starting point.

For B:

at time ti, B moved from Xo to 2Xo, displacement is 2Xo.

at time t2 a moves with speed V in the opposite position so he'll be back to his starting point, hence, his new position will be at Xo. A's displacement is 0 from his starting point.

Answer 2

The average velocity of car A between t₁ and t₂ is greater than the average velocity of car B between t₁ and t₂.

What is average velocity?

The average velocity is the ratio of the total displacement traveled to the total time taken by the body. Its unit is m/sec.

The given data in the problem is;

t₁ is the time at position 1

xo is the position of car A and car B

v is the speed of cars A and B

t₂ is the time at position 2

3v is the speed of car A

v is the speed of car B

The average velocity is the ratio of the total distance traveled to the total time taken by the body. Its unit is m/sec.

At time t₁ A moved from Xo to 2Xo, resulting in a displacement of 2Xo.

At time t2, A travels at a speed of 3V, therefore his new position is 3Xo, as opposed to 2Xo, which is at 5Xo. A's displacement from the starting location is 5Xo.

B moved from Xo to 2Xo at time ti, and the displacement is 2Xo.

At time t₂ proceeds with speed V in the opposite direction, returning to his starting location, and therefore his new position is at Xo. A's the displacement from his starting location is zero.

$\rm V_{avg}= \frac{total \ displacement}{total \ time}$

The displacement of car A between t₁ and t₂ is greater than the displacement of car B between t₁ and t₂.

Hence the average velocity of car A between t₁ and t₂ is greater than the average velocity of car B between t₁ and t₂.

To learn more about the average velocity refer to the link;

https://brainly.com/question/862972