Answer :
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
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:
[tex]v_{avg} =\frac{x_{f} -x_{o}}{t_{f} - t_{o} } (1)[/tex]
- 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:
[tex]v_{avg} =\frac{x_{f} -x_{o}}{t_{f} - t_{o} } =\frac{98 m -12 m}{10.0 s - 0 s } = \frac{86m}{10.0s} = 8.6 m/s (2)[/tex]