Question

an object on a number like moved from x = 12 m to x = 124 m and moved back to x = 98 m. the time interval for all the motion was 10 s. what was the average velocity of the object?

Answer 1

Complete Question:

an object on a number line moved from x = 12 m to x = 124 m and moved back to x = 98 m. the time interval for all the motion was 10 s. what was the average velocity of the object?

Answer:

v = 8.6 m/s

Step-by-step explanation:

• By definition, the average velocity, is just the rate of change of the position (denoted by the x-coordinate in this case) with respect to time, as follows:

`v_(avg) =(x_(f) -x_(o))/(t_(f) - t_(o) ) (1)`

• This means that the average velocity depends only on the final and initial positions, independent from the intermediate points between them.
• As we can choose freely our origin in time, we choose the initial time to be zero.
• At that time, the x-coordinate is x=12 m, so x₀ = 12 m.
• When t= 10.0 s, x is 98 m, so xf = 98 m.
• Replacing in (1) we get:

`v_(avg) =(x_(f) -x_(o))/(t_(f) - t_(o) ) =(98 m -12 m)/(10.0 s - 0 s ) = (86m)/(10.0s) = 8.6 m/s (2)`