2.5 Velocity and Acceleration

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1 82 CHAPTER 2. VECTOR FUNCTIONS 2.5 Velocity and Acceleration In this section, we study the motion of an object alon a space curve. In other words, as the object moves with time, its trajectory follows a space curve. We will see how the notions of tanent vector, normal vector and curvature apply to describe the motion of the object Definitions Definition 124 (Velocity and Acceleration) Consider an object movin alon C, a smooth curve, twice differentiable, with position vector r (t). 1. The velocity of the object, denoted v (t) is defined to be r (t) (2.14) 2. The acceleration of the object, denoted a (t) is defined to be a (t) = r (t) = v (t) (2.15) 3. The speed of the object is v (t). Example 125 (Findin the velocity and acceleration of a movin object) An object is movin alon the curve r (t) = t, t 3, 3t for t 0. Find v (t), a (t) and sketch the trajectory of the object as well as the velocity and acceleration when t =1. Velocity: Acceleration: 1, 3t 2, 3 a (t) =0, 6t, 0 Sketch: When t =1,wehave v (1) = 1, 3, 3 and a (1) = 0, 6, 0 The trajectory of the object as well as the velocity and acceleration at t =1 are shown in fiure 2.8. In many applications, we do not know the position function. Instead, we know the acceleration and we must find the velocity and position function. The next example illustrates this.

2 2.5. VELOCITY AND ACCELERATION 83 Fiure 2.8: Motion of an object alon r (t) = t, t 3, 3t Example 126 (Findin a Position Function by Interation) A movin particle starts at position r (0) = 1, 0, 0 with initial velocity v (0) = 1, 1, 1. Its acceleration is a (t) =4t, 6t, 1. Find its velocity and position function at time t. Velocity: Since a (t) = v (t), it follows that a (t) dt Thus 2t 2 + C 1, 3t 2 + C 2,t+ C 3 We find the constants by usin the initial condition. v (0) = 1, 1, 1 = C1,C 2,C 3 Thus, C 1 =1, C 2 = 1 and C 3 =1. It follows that 2t 2 +1, 3t 2 1,t+1 Position Function: Since r (t), it follows that r (t) = v (t) dt = 2t t + C 1,t 3 t + C 2, t2 2 + t + C 3

3 84 CHAPTER 2. VECTOR FUNCTIONS We find the constants by usin the initial condition. r (0) = 1, 0, 0 = C1,C 2,C 3 Thus, C 1 =1and C 2 = C 3 =0. It follows that 2t 3 r (t) = 3 + t +1,t3 t, t2 2 + t Remark 127 Remark 128 In eneral, we can recover velocity by interation when acceleration is known. Similarly, we can recover position by interation when velocity is known. More precisely, if a (t 0 ) and v (t 0 ) are known, then t v (t0 )+ a (u) du t 0 Similarly, if v (t) and r (t 0 ) are known, then t r (t) = r (t0 )+ v (u) du t 0 Remark 129 When studyin the motion of an object, the time at which we start trackin the object is usually set to 0. The acceleration at time t =0is called initial acceleration and denoted a 0. Similarly, the velocity at time t =0is called initial velocity and usually denoted v 0. The position at time t =0is called initial position and usually denoted r 0. The term initial means at time t = Projectile Motion If you wondered how we can know the acceleration and not the other quantities, this section will hopefully answer your questions. We bein with a little bit of physics. Newton s second law of motion states that if at time t aforce F (t) acts on an object of mass m, this action will produce an acceleration a (t) of the object satisfyin F (t) =m a (t) Thus, if we know the force actin on the object, we can find the acceleration. We will then be in the situation of the previous example. We illustrate this with a classical problem in physics, the problem of findin the trajectory of an object bein thrown in the air and subject to the laws of physics. Example 130 A projectile is fired with an anle of elevation α and initial velocity v 0 as shown in fiure 2.9. If we nelect the air resistance, the only force actin on the object is ravity. 1. Find the position of the object r (t).

4 2.5. VELOCITY AND ACCELERATION Express the rane d in terms of α. 3. Find the value of α which maximizes the rane d. Answer to 1. We bein by settin the axes so that y is the vertical direction and x the horizontal direction. In addition, the motion of the projectile beins at the oriin. The only force the object is subject to is ravity. Since the force of ravity acts downward, we have a (t) =0, where =9.81m/s 2. We can now compute v (t). a (t) dt = C 1, t + C 2 Let v 0 = v 0x,v 0y. Then, we have v (0) = v0x,v 0y = C 1,C 2 Thus v0x, t + v 0y Elementary trionometry tells us that v 0x = v 0 cos α v 0y = v 0 sin α Thus v 0 cos α, t + v 0 sin α Now, we can compute r (t). r (t) = v (t) dt = ( v 0 cos α) t + C 1, 12 t2 +( v 0 sin α) t + C 2 Since r (0) = 0, 0, bothc 1 and C 2 are 0 hence r (t) = ( v 0 cos α) t, 1 2 t2 +( v 0 sin α) t So, we can see that the motion in the x direction and in the y direction follow different laws. This can be made more obvious by writin the parametric equations of the trajectory of the object. { x =( v 0 cos α) t y = 1 2 t2 +( v 0 sin α) t

5 86 CHAPTER 2. VECTOR FUNCTIONS Fiure 2.9: Trajectory of a Projectile Answer to 2. d, the rane, is the horizontal distance correspondin to y =0. We can find d by findin the value of t for which y =0and pluin this value in the equation for x. y = t2 +( v 0 sin α) t =0 ( t 1 ) 2 t + v 0 sin α =0 This happens when either t =0or t = 2 v 0 sin α. t =0corresponds to the initial position. The value of t we want is t = 2 v 0 sin α. This value ives us d = ( v 0 cos α) 2 v 0 sin α d = v 0 2 sin 2α Answer to 3. We see that d is maximum when sin 2α =1that is when α = π 4. Make sure you have read, studied and understood what was done above before attemptin the problems.

6 2.5. VELOCITY AND ACCELERATION Problems Do # 3, 5, 7, 9, 11, 13, 15, 17, 21, 23, 25 on pae 725.

2.2 Differentiation and Integration of Vector-Valued Functions

2.2 Differentiation and Integration of Vector-Valued Functions .. DIFFERENTIATION AND INTEGRATION OF VECTOR-VALUED FUNCTIONS133. Differentiation and Interation of Vector-Valued Functions Simply put, we differentiate and interate vector functions by differentiatin