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(f) Because of what we learned in parts (d) and (e), we need to calculate the distances traveled during the time intervals [0, 1], [1, 3], and [3, 5] separately. The distance traveled in the first 1 second is |(1) – f(0)| = O -D = 4 m. From t = 1 to t = 3 the distance traveled is |F(3) – f(1)| = m. From t = 3 to t = 5 the distance traveled is |F(5) – f(3)| = = 20 m. The total distance is 4 + + 20 = (g) The acceleration is the derivative of the velocity function: d?s dt? dv = 6t – 12 dt a(t) a(4) 6(4) – 12 = m/s². %3D

(f) Because of what we learned in parts (d) and (e), we need to calculate the distances traveled during the time intervals [0, 1], [1, 3], and [3, 5] separately. The distance traveled in the first 1 second is |(1) – f(0)| = O -D = 4 m. From t = 1 to t = 3 the distance traveled is |F(3) – f(1)| = m. From t = 3 to t = 5 the distance traveled is |F(5) – f(3)| = = 20 m. The total distance is 4 + + 20 = (g) The acceleration is the derivative of the velocity function: d?s dt? dv = 6t – 12 dt a(t) a(4) 6(4) – 12 = m/s². %3D

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section11.4: Plane Curves And Parametric Equations
Problem 40E
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(e) Using the information from part (d) we make a schematic sketch in the figure of the motion of the particle back and forth along a line (the s-axis). (This sketch is to the left.) (f) Because of what we learned in parts (d) and (e), we need to calculate the distances traveled during the time intervals [0, 1], [1, 3], and [3, 5] separately. The distance traveled in the first 1 second is |(1) – f(0)| = 4 m. From t = 1 to t = 3 the distance traveled is |F(3) – f(1)| = m. From t = 3 to t = 5 the distance traveled is |F(5) – f(3)| = 20 m. The total distance is 4 + + 20 = m (g) The acceleration is the derivative of the velocity function: d2s dt? dv a(t) = 6t - 12 dt a(4) = 6(4) – 12 = m/s?. (h) The figure shows graphs of s, v, and a.
Transcribed Image Text:(e) Using the information from part (d) we make a schematic sketch in the figure of the motion of the particle back and forth along a line (the s-axis). (This sketch is to the left.) (f) Because of what we learned in parts (d) and (e), we need to calculate the distances traveled during the time intervals [0, 1], [1, 3], and [3, 5] separately. The distance traveled in the first 1 second is |(1) – f(0)| = 4 m. From t = 1 to t = 3 the distance traveled is |F(3) – f(1)| = m. From t = 3 to t = 5 the distance traveled is |F(5) – f(3)| = 20 m. The total distance is 4 + + 20 = m (g) The acceleration is the derivative of the velocity function: d2s dt? dv a(t) = 6t - 12 dt a(4) = 6(4) – 12 = m/s?. (h) The figure shows graphs of s, v, and a.
t = 3 S = 0 EXAMPLE 9 The position of a particle is given by the equation f(t) = t3 - 6t2 + 9t S = where t is measured in seconds and s in meters. (a) Find the velocity at time t. (b) What is the velocity after 2 s? After 4 s? (c) When is the particle at rest? t = 0 S = 0 t = 1 s = 4 (d) When is the particle moving forward (that is, in the positive direction)? (e) Draw a diagram to represent the motion of the particle. y 25 (f) Find the total distance traveled by the particle during the first five seconds. (g) Find the acceleration at time t and after 4 s. (h) Graph the position, velocity, and acceleration functions for 0 st< 5. SOLUTION (a) The velocity function is the derivative of the position function. a S = f(t) = t3 - 6t2 + 9t S ds 3t dt v(t) 12t + 9 (b) The velocity after 2 s means the instantaneous velocity when t = 2, that is ds v(2) : 3(2)2 – 12(2) + 9 = |-3 m/s. dt t = 2 -12 Video Example The velocity after 4 s is v(4) = 3(4)2 - 12(4) + 9 = |9 m/s. (c) The particle is at rest when v(t) = 0 that is, 3t? – 12t + 9 = 3(t2 – 4t + 3) = 3(t – 1)(t – 3) = |0 and this is true when t = 1 (smaller t-value) or t = 3 (larger t-value). Thus the particle is at rest after 1 s (smaller t-value) and after 3 S (larger t-value). (d) The particle moves in the positive direction when v(t) 0, that is, 3t2 - 12t + 9 3(t – 1)(t – 3) 0. This inequality is true when both factors are positive t > 3 or when both factors are negative (t - ). <1 Thus the particle moves in the positive direction in the time intervals t < 1 and t > 3 It moves backward (in the negative direction) when 1 <t < 3
Transcribed Image Text:t = 3 S = 0 EXAMPLE 9 The position of a particle is given by the equation f(t) = t3 - 6t2 + 9t S = where t is measured in seconds and s in meters. (a) Find the velocity at time t. (b) What is the velocity after 2 s? After 4 s? (c) When is the particle at rest? t = 0 S = 0 t = 1 s = 4 (d) When is the particle moving forward (that is, in the positive direction)? (e) Draw a diagram to represent the motion of the particle. y 25 (f) Find the total distance traveled by the particle during the first five seconds. (g) Find the acceleration at time t and after 4 s. (h) Graph the position, velocity, and acceleration functions for 0 st< 5. SOLUTION (a) The velocity function is the derivative of the position function. a S = f(t) = t3 - 6t2 + 9t S ds 3t dt v(t) 12t + 9 (b) The velocity after 2 s means the instantaneous velocity when t = 2, that is ds v(2) : 3(2)2 – 12(2) + 9 = |-3 m/s. dt t = 2 -12 Video Example The velocity after 4 s is v(4) = 3(4)2 - 12(4) + 9 = |9 m/s. (c) The particle is at rest when v(t) = 0 that is, 3t? – 12t + 9 = 3(t2 – 4t + 3) = 3(t – 1)(t – 3) = |0 and this is true when t = 1 (smaller t-value) or t = 3 (larger t-value). Thus the particle is at rest after 1 s (smaller t-value) and after 3 S (larger t-value). (d) The particle moves in the positive direction when v(t) 0, that is, 3t2 - 12t + 9 3(t – 1)(t – 3) 0. This inequality is true when both factors are positive t > 3 or when both factors are negative (t - ). <1 Thus the particle moves in the positive direction in the time intervals t < 1 and t > 3 It moves backward (in the negative direction) when 1 <t < 3
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