He can hold the rifle at an angle of about 7 ∘ with the horizontal and hit his target 0.07s after firing, or he can hold his rifle almost straight up, with an angle of 89.8 ∘, where he’ll hit his target about 22s later. Using cos ⁡ θ = 25 / ( 350 ⁢ t ), we can solve for θ: That is, we can apply the quadratic formula to find t 2, then solve for t itself. Using the Pythagorean Identity cos 2 ⁡ θ + sin 2 ⁡ θ = 1, we have We start by solving each equation for cos ⁡ θ and sin ⁡ θ, respectively.Ĭos ⁡ θ = 25 350 ⁢ t and sin ⁡ θ = 3 + 16 ⁢ t 2 350 ⁢ t.

That is, we want to find θ and t such that We need to find θ so that r → ⁢ ( t ) = ⟨ 25, 6 ⟩ for some value of t. SolutionThe position function for the path of Alex’s bb is Instantaneous acceleration is a vector in two or three dimensions. The instantaneous acceleration is produced by a change in velocity taken over a very short (infinitesimal) time period. R → ⁢ ( t ) = ⟨ ( v 0 ⁢ cos ⁡ θ ) ⁢ t + x 0, - 1 2 ⁢ g ⁢ t 2 + ( v 0 ⁢ sin ⁡ θ ) ⁢ t + y 0 ⟩. In two and three dimensions, the acceleration vector can have an arbitrary direction and does not necessarily point along a given component of the velocity. Knowing r → ⁢ ( 0 ) = ⟨ x 0, y 0 ⟩, we conclude C → = ⟨ x 0, y 0 ⟩ and We integrate once more to find r → ⁢ ( t ): Knowing v → ⁢ ( 0 ) = v 0 ⁢ ⟨ cos ⁡ θ, sin ⁡ θ ⟩, we have C → = v 0 ⁢ ⟨ cos ⁡ θ, sin ⁡ θ ⟩ and so † † margin: Note: In this text we use g = 32ft/s 2 when using Imperial units, and g = 9.8m/s 2 when using SI units. Since the acceleration of the object is known, namely a → ⁢ ( t ) = ⟨ 0, - g ⟩, where g is the gravitational constant, we can find r → ⁢ ( t ) knowing our two initial conditions. This means that every 1/2 second after t = 0.016s the boy can release the string (since the ball makes 2 revolutions per second, he has two chances each second to release the ball). 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.

Direction of acceleration vector 2d