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Position, Velocity and Acceleration 3.2 Position, Velocity and Acceleration All fall under rectilinear motion Motion along a straight line We are normally given a function relating the position of a moving object with respect to time. Velocity is the derivative of position Acceleration is the derivative of velocity Position, Velocity and Acceleration Position Velocity v(t) or s’(t) Acceleration s(t) or x(t) a(t) or v’(t) or s’’(t) Speed is the absolute value of velocity Example If the position of a particle at time t is given by the equation below, find the velocity and acceleration of the particle at time, t = 5. s(t ) t 3 11t 2 24t v(t ) s' (t ) 3t 2 22t 24 v(5) 3(5) 22(5) 24 11 2 a(t ) v' (t ) s' ' (t ) 6t 22 a(5) 6(5) 22 8 Position, Velocity and Acceleration When velocity is negative, the particle is moving to the left or backwards When velocity is positive, the particle is moving to the right or forwards When velocity and acceleration have the same sign, the speed is increasing When velocity and acceleration have opposite signs, the speed is decreasing. When velocity = 0 and acceleration does not, the particle is momentarily stopped and changing direction. Example If the position of a particle is given below, find the point at which the particle changes direction. s(t ) t 12t 36t 18 3 2 v(t ) 3t 2 24t 36 a(t ) 6t 24 a(6) 12 a(2) 12 Changes direction when velocity = 0 and acceleration does not 0 3t 2 24t 36 0 t 2 8t 12 0 (t 6)(t 2) t 6, 2 Example Using the previous function, find the interval of time during which the particle is slowing down. V(t) = 0 at 2 and 6, a(t) = 0 at 4 s(t ) t 12t 36t 18 3 2 v(t ) 3t 2 24t 36 a(t ) 6t 24 0 Particle is slowing down when, 0<t<2 4<t<6 t v(t) a(t) 4 2 + - - 6 + + + Example When velocity = 0 When does this occur? How far does a particle travel between the eighth and tenth seconds if its position is given by: s(t ) t 2 6t To find the total distance we must find if the particle changes directions at any time in the interval v(t ) 2t 6 0 2t 6 3t The object may travel forward then backwards, thus s(10) – s(8) is really only the 3 displacement is not in our interval it will not theso total not affect our problem! distance! s(10) s(8) 40 16 24 Example How far does a particle travel between zero and four seconds if its position is given by: s(t ) t 4 8t 2 v(t ) 4t 3 16t s(2) s(0) s(4) s(2) 16 32 12816 v(t ) 4t (t 2 4) 0 4t (t 2)(t 2) 160 t 0, 2, 2 Divide into intervals; 02 and 24 At any time t, the position of a particle moving along an axis is: s(t ) t 3 6t 2 9t A. Find the body’s acceleration each time the velocity is zero v(t ) 3t 12t 9 s(1) 0s(0t)2 4s(t2) 3 s(1) 2 C. Find the total distance traveled by the body from t = 0 to t = 2 4 00 (2t 34)(t 1) 6 t 3, 1 a(t ) 6t 12 a(1Velocity ) 6= 0 at 1! a(3) 6 Divide into intervals; 01 and 12 B. Find the body’s speed each time the acceleration is zero a(t ) 6t 12 0 6t 12 t2 v(2) 12 24 9 3 At any time t, the position of a particle moving along an 3 axis is: t 2 s(t ) 3 2t 3t A. When is the body moving forward? backwards? v(t ) t 2 4t 3 v(t) 0 (t 3)(t 1) t 3, 1 1 3 + - + Forward (0, 1) and (3, ∞) Backwards from (1, 3) B. When is the velocity increasing? decreasing? a(t ) 2t 4 Velocity increasing: (1, 2) and (3, t ∞)2 Velocity decreasing: (0, 1) and (2, 3) 0 t v(t) a(t) 1 + - 2 - 3 + + +