This line segment is tangent to the circle, which means it is also perpendicular to r → ( t 0 ) itself, so their dot product is 0. As shown in Figure 12.3.4, the vector from the release point to the tree is ⟨ 0, 10 ⟩ - r → ( t 0 ). We choose one that is relatively simple computationally. There are many ways to find this time value. We want to find t 0 so that this line contains the point ( 0, 10 ) (since the tree is 10ft directly in front of the boy). The ball will be at r → ( t 0 ), traveling in the direction of v → ( t 0 ). Let t = t 0 be the time when the boy lets go of the string. When the boy releases the string, the string no longer applies a force to the ball, meaning acceleration is 0 → and the ball can now move in a straight line in the direction of v → ( t ). † † margin: 2ft ⟨ 0, 10 ⟩ - r → ( t 0 ) Figure 12.3.4: Modeling the flight of a ball in Example 12.3.3. When velocity is changing rapidly, the acceleration must be “large.” A ball whirling quickly is rapidly changing direction/velocity. The magnitude of the acceleration is related to the speed at which the ball is traveling. In what direction is this force/acceleration being applied? In the direction of the string, towards the boy’s hand. This is not acceleration in the sense of “it travels faster ” rather, this acceleration is changing the velocity of the ball. The string applies a force to the ball, affecting it’s motion: the force accelerates the ball. Why does the ball in our example move in a circle? It is attached to the boy’s hand by a string. A moving ball “wants” to travel in a straight line. Thus force and acceleration are closely related.
Recall the classic physics equation, “Force = mass × acceleration.” A force acting on a mass induces acceleration (i.e., the mass moves). Note how a → ( t ) is parallel to r → ( t ), but has a different magnitude and points in the opposite direction. To find a → ( t ), we differentiate r → ( t ) twice.
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