Imagine a ball is moving on the following horizontal line.
. . . . . . . . . . . . . . . . . . . O. . . . . . . . . . . . . . . . . .
Take right as positive. O is the starting point of the ball. Denote the ball by o.
. . . . . . . . . . . . . . . . . . . O. . . . . . . ... . . o . . . . . .
Assume the ball is moving to the right. It has positive displacement since it is on the right of O, and positive velocity since its positive displacement is increasing.
.ñ
. . . . . . . . . . . . . . . . . . . O. . . . o . . . . . . . . . . . . .
Now the ball is returning to O. It still has positive displacement since its current position is still on the right of O. However, its velocity is negative since its positive displacement is decreasing and the direction of the velocity vector points left, which is the negative side.
By now you should be able to come up with a scenario where the ball has negative displacement and positive velocity.
You can observe the same phenomenon in daily life. Say, as a stretched spring bounces to its starting position, if we let the returning direction be positive, the string has negative displacement since it is on the negative direction, but has positive velocity. Bungee jump can also used to illustrate the phenomenon.
Inclined plane is a simple machine that is a straight slanted surface.
Answer:
270 m/s²
Explanation:
Given:
α = 150 rad/s²
ω = 12.0 rad/s
r = 1.30 m
Find:
a
The acceleration will have two components: a radial component and a tangential component.
The tangential component is:
at = αr
at = (150 rad/s²)(1.30 m)
at = 195 m/s²
The radial component is:
ar = v² / r
ar = ω² r
ar = (12.0 rad/s)² (1.30 m)
ar = 187.2 m/s²
So the magnitude of the total acceleration is:
a² = at² + ar²
a² = (195 m/s²)² + (187.2 m/s²)²
a = 270 m/s²
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