15. A jogger runs down a straight stretch of road with an average velocity of 4.00 m/s for 2.00 min. What is his final position, taking his initial position to be zero?
Answers
Answer:
Let us imagine that a jogger is running on a track. Notice that the runner runs more slowly on the average during the second time interval as if she is tiring.
Let us use the two separate portions of motion and find out the displacement for each portion.
Velocity = rate of change of displacement/ rate of change of time.
So, Rate of change of displacement = Velocity times rate of change of time.
Therefore, the rate of change of displacement from position 1
= 5.00m/s×4.00min×1min60s=1.20×103m.
The rate of change of displacement from position 2 = 4.00m/s×3.00min×1min60s=.720×103m.
We add up both these displacements to find the total displacement of 1.92×103m.
Hence, the magnitude of the final displacement from her initial position is 1.92×103m.
Answer:
480m
Explanation:
1. Identify the knowns. v–=4.00 m/s,Δt=2.00 min, and x0=0 m.
2. Enter the known values into the equation.
2.32
x=x0+v–t=0+(4.00 m/s)(120 s)=480 m.