Why Is Average Velocity Calculated as the Mean of Initial and Final Velocities?

[xzibition8612]

Homework Statement

A vehicle accelerates from 2 m/s to 26 m/s in 12 seconds. Determine the acceleration. Also find the average velocity and the distance travelled.

Homework Equations

Average velocity = Total distance/ time taken

The Attempt at a Solution

Acceleration = (26-2)/12 = 2m/s²
Average velocity = ?

Ok the answer for average velocity is = (26+2)/2 = 12.
Total distance = 14*12 =168m.

My question is how is the average velocity computed like this?? WHy 26+2? And why divided by 2? Isn't the total time taken 12 seconds? SHouldn't it be divided by 12? ANy help would be Appreciated.

 

[SammyS]

Homework Statement

A vehicle accelerates from 2 m/s to 26 m/s in 12 seconds. Determine the acceleration. Also find the average velocity and the distance travelled.

Homework Equations

Average velocity = Total distance/ time taken

The Attempt at a Solution

Acceleration = (26-2)/12 = 2m/s²
Average velocity = ?

Ok the answer for average velocity is = (26+2)/2 = [STRIKE]12[/STRIKE] 14 m/s.
Total distance = 14*12 =168m.

My question is how is the average velocity computed like this?? WHy 26+2? And why divided by 2? Isn't the total time taken 12 seconds? SHouldn't it be divided by 12? ANy help would be Appreciated.

The average velocity can be computed like that only if acceleration is uniform (constant). Also, if the acceleration were not constant, all the could be computed in this problem is average acceleration.

 

[xzibition8612]

but why divided by 2? So if acceleration is constant its always divide by 2?

 

[xzibition8612]

damn I'm so goddamn stupid not to understand this