Chapter 12, Problem 47P

1- A particle moves along the x axis. Its position varies with time according to the expression x=-41+ 2f where x is in meters and t is in seconds. The position-time graph for this motion is shown in Figure. Note that the particle moves in the negative x direction for the first second of motion, is momentarily at rest at the moment 1 = 1 s, and moves in the positive x direction at times /> 1 s. (A) Determine the displacement of the particle in the time intervals t=0 to 1=1 s and t= 1 s to / = 3 s. B) Calculate the average velocity during these two-time intervals C) determine the instantaneous velocity and instantaneous speed at t= 0.5 s. x(m) Slope = 4 m/s 10 00 8 9 10 O Slope =-2 m/s 1 2 015 t(s)

A particle's position is given by s = (2.17 + 4.80t2 0.100t*)m where t is in seconds and s is in meters. Find - – the particles position and acceleration at the instant when the particle has zero velocity. You may use mode + equation in your calculator to find the roots.

A train started moving from rest with a constant acceleration, and at one time it was traveling at a speed of sec/m30 and after a distance of 160m it was traveling at a speed of sec/m50. Find: 1: the wheel. 2: The time it takes to reach its velocity sec/m30.

The x-coordinate of a particle in curvilinear motion is given by x = 5.1t3 - 5.3t where x is in feet and t is in seconds. The y-component of acceleration in feet per second squared is given by ay = 1.5t. If the particle has y-components y = 0 and vy = 4.5 ft/sec when t = 0, find the magnitudes of the velocity v and acceleration a when t = 2.8 sec. Sketch the path for the first 2.8 seconds of motion, and show the velocity and acceleration vectors for t = 2.8 sec.

The x-coordinate of a particle in curvilinear motion is given by x = 5.3t³ - 5.1t where x is in feet and t is in seconds. The y-component of acceleration in feet per second squared is given by ay = 3.8t. If the particle has y-components y = 0 and vy = 3.0 ft/sec when t = 0, find the magnitudes of the velocity v and acceleration a when t = 4.5 sec. Sketch the path for the first 4.5 seconds of motion, and show the velocity and acceleration vectors for t = 4.5 sec. Answers: V = i ft/sec i ft/sec² a =

TACTICS BOX 1.2: Finding the acceleration vector. To find the acceleration as the velocity changes from ₁ to vƒ : (Figure 1) 27 1. Draw the velocity vectors V₁ and vf and with their tails together. 2 V 2. Draw the vector from the tip of V; to the tip of Vf. This is A because of = vi + Av. 3. Return to the original motion diagram. Draw a vector at the middle dot in the direction of Av; label it a. This is the average acceleration at the midpoint between ₁ and Vf. Vf 15

The x-coordinate of a particle in curvilinear motion is given by x = 2t² - 3t where x is in meters and t is in seconds. Also, the particle has an acceleration in the y-direction given by a, 8- sec. If the velocity of the particle in the y-direction is 3 m/sec when t=0, calculate the magnitudes of the velocity v and acceleration a of the particle when t=2 sec. and the angles these vectors make with the x-axis at the same instant.

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The y-coordinate of a particle in curvilinear motion is given by y = 2.0t3 - 8.1t, where y is in inches and t is in seconds. Also, the particle has an acceleration in the x-direction given by ax = 2.4t in./sec2. If the velocity of the particle in the x-direction is 5.0 in./sec when t = 0, calculate the magnitudes of the velocity v and acceleration a of the particle when t = 2.4 sec. Construct v and a in your solution.Answers: When t = 2.4 sec,