i REALLY don't get this problem. PLEASE HELP!!!
the position of a runner during a 3 second time interval the runners position changes from x1 = 50m to x2 = 30.5m what was the runners average velocity?
find the average speed as well as the average velocity.
The velocity is both magnitude and direction. Let's consider velocity in the +x direction positive.
Then the average velocity would be
V = (x2-x1)/t = (30.5-50)/3 = -6.5 m/s
which is negative.
The speed however is the magnitude of that vector or +6.5 m/s
If you have had the "unit vector" in your physics subject so far, then a unit vector in the x direction would be i, in the y direction j and in the z direction k
In this case the velocity is -6.5 in the x direction so you would write the velocity vector V as
V = -6.5 i
or V = -6.5 i + 0 j + 0 k
thanks a lot!!!!
To find the average velocity of the runner, we need to calculate the displacement and divide it by the time interval.
Displacement (Δx) is the change in position of an object. In this case, it is the difference between the initial position (x1) and the final position (x2).
Δx = x2 - x1
Δx = 30.5m - 50m
Δx = -19.5m
The negative sign indicates that the runner moved in the opposite direction (towards the negative direction on the coordinate system).
Now, we need to divide the displacement by the time interval to find the average velocity.
Time interval (Δt) is given as 3 seconds.
Average velocity (v_avg) = Δx/Δt
v_avg = -19.5m/3s
v_avg ≈ -6.5 m/s
Therefore, the average velocity of the runner is approximately -6.5 m/s.
Now, to find the average speed, we need to calculate the total distance traveled and divide it by the time interval.
Total distance traveled (d) is the sum of the magnitudes of the initial and final positions.
d = |x2| + |x1|
d = |30.5m| + |50m|
d = 30.5m + 50m
d = 80.5m
Again, divide the total distance traveled by the time interval to find the average speed.
Average speed (s_avg) = d/Δt
s_avg = 80.5m/3s
s_avg ≈ 26.83 m/s
Therefore, the average speed of the runner is approximately 26.83 m/s.