A motorboat accelerates uniformly from a ve-
locity of 6.2 m/s to the west to a velocity of
1.7 m/s to the west.
If its acceleration was 3.4 m/s2 to the east,
how far did it travel during the acceleration?
Answer in units of m
Your umpteen posts have been removed.
Once you include YOUR THOUGHTS about solving your problems, please re-post, and someone here will be happy to comment.
The knight 'twas darkith
To find the distance traveled during acceleration, we can use the formula:
distance = (final velocity^2 - initial velocity^2) / (2 * acceleration)
In this case, the initial velocity (u) is 6.2 m/s to the west, the final velocity (v) is 1.7 m/s to the west, and the acceleration (a) is 3.4 m/s^2 to the east.
First, let's convert the velocities to their magnitudes by ignoring the direction:
initial velocity magnitude (|u|) = 6.2 m/s
final velocity magnitude (|v|) = 1.7 m/s
Now, we can calculate the distance using the formula:
distance = (|v|^2 - |u|^2) / (2 * |a|)
Substituting the given values:
distance = (1.7^2 - 6.2^2) / (2 * 3.4)
Calculating the numerator:
distance = (2.89 - 38.44) / (2 * 3.4)
distance = (-35.55) / (6.8)
Finally, calculating the distance:
distance = -5.23 m
The negative sign indicates that the acceleration and initial velocities were in opposite directions, resulting in a decrease in distance traveled.
Therefore, the motorboat traveled a distance of 5.23 m during the acceleration.