1. The position function of a particle is given by s(t) = t³ – 6t²+9t – 60, t > 0 is given in seconds.a. Where is the particle located at the second instant it changes direction?b. What is the distance travelled by the particle after 5 seconds?

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Answered: 1. The position function of a particle…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Suppose the velocity of an object moving along a line is positive. Are position, displacement, and distance traveled equal? Explain. Choose the correct answer below. O A. Yes, if the velocity is positive then the displacement, distance traveled, and position of the object will be given b by v(t) dt. a O B. No, the displacement and position of the object will be equal but since the initial position is not given, the distance traveled by the object may not be equal to the position and the displacement of the object. O C. Yes, if the velocity is positive then the displacement, distance traveled, and position of the object will be given by v'(t). O D. No, the displacement and distance traveled by the object will be equal but since the initial position is not given, the position of the object may not be equal to the distance traveled and the displacement of the object.
Suppose the height of a ball can be measured by S(t) = (t + 2)² – 5, where the height is measured in feet and time in seconds. Find the average velocity of the ball fromt = 2 to t = 4. Select one: O 4 ft/sec 8 ft/sec 25 ft/sec 10 ft/sec 20 ft/sec
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13) The integer n = a) If n = 3 (mod 9), then what is x ? b) If n = 3 (mod 11), then what is x ? Hint for #13: 12700140x3243243 is missing the digit x. Consider the number 1234. By writing each digit in scientific notation, we have 1234 1 x 10³ +2× 10² +3 × 10¹ + 4 × 10⁰ Then notice that 10 = 1 (mod 9). When we reduce mod 9, we obtain 1234 1 x 10³ +2× 10² +3 x 10¹ + 4 x 10⁰ = = 1x 1³ + 2 x 1² + 3 × 1¹ + 4 × 1⁰ = 1+ 2+ 3+ 4 = 10 = 1 (mod 9). Since 10 = -1 (mod 11), when we reduce mod 11, we obtain 1234 1 x 10³ +2× 10² +3 x 10¹ + 4 x 10⁰ = = 1x (-1)³ +2× (-1)² +3 × (-1)¹ +4× (-1)⁰ = -1 + 2-3 + 4 = 2 (mod 11).

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1. The position function of a particle is given by s(t) = t³ – 6t²+9t – 60, t > 0 is given in seconds. a. Where is the particle located at the second instant it changes direction? b. What is the distance travelled by the particle after 5 seconds?