At time t = 14 s, the velocity of a particle moving in the x-y plane is v = 0.13i + 2.60j m/s. By time t =14.09 s, its velocity has become-0.06i + 2.50j m/s. Determine the magnitude aay of its average acceleration during this interval and the angle e made by the averageacceleration with the positive x-axis. The angle e is measured counterclockwise from the positive x-axis.Answers:dav =im/s2e =

Given data: V14s=0.13i+2.6j m/sV14.09s=-0.06i+2.5j m/s Need to determine the acceleration of the and…

Answered: At time t= 14 s, the velocity of a…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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8.2 The velocity of a particle moving in the x-y plane is given by (5.60i + 4.27j) m/s at time t = 6.33 s. Its average acceleration during the next 0.024 s is (6.5i + 5.4j) m/s2. Determine the velocity v of the particle at t = 6.354 s and the angle θ between the average-acceleration vector and the velocity vector at t = 6.354 s.
The velocity of a particle along the s-axis is given by v = 10s5/4 where s is in millimeters and v is in millimeters per second. Determine the acceleration when s is 3.6 millimeters. Answer: a = mm/s²
A motorcycle cage has a curved path in the x-z plane as shown. The cage path has different curved profile, and between point o and A the path is best described as z = f(x) = ax², where a = 0.09 and b = 1.23. The motorcyclist at point M is moving at a velocity of 12.6 m/s which is increasing at a rate of 1.9 m/s?. If the horizontal position of the motorcyclist at point M is xM = 11.9 m, determine (i) the direction of the motorcyclisť's velocity. (ii) the radius of the curvature of the cage's path at point M, (iii) the normal acceleration at point M, (iv) the magnitude and direction of the motorcyclist's acceleration at point M. A (z = ax
A particle which moves with curvilinear motion has coordinates in millimeters which vary with the time t in seconds according to x = 3t² — 4tand y = 4t² - ²/t³. Determine the magnitudes of the velocity v and acceleration a and the angles which these vectors make with the x-axis when t = 2 s. Ans. v=14.42 mm/s, a = 7.2 mm/s², 0, = 56.3⁰, 0a = 33.7⁰
The velocity of a particle moving in the x-y plane is given by (4.32i + 5.97j) m/s at time t = 5.50 s. Its average acceleration during the next 0.029 s is (7.1i + 6.4j) m/s2. Determine the velocity v of the particle at t = 5.529 s and the angle θ between the average-acceleration vector and the velocity vector at t = 5.529 s.
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)
The velocity of a particle moving in the x-y plane is given by (5.00i + 7.38j) m/s at time t = 3.87 s. Its average acceleration during the next 0.014 s is (2.9i+2.6j) m/s². Determine the velocity v of the particle at t = 3.884 s and the angle between the average- acceleration vector and the velocity vector at t = 3.884 s. Answers: v = ( i e = i i+ i j) m/s
Question 7 A particle which moves with curvilinear motion has coordinates in millimeters which vary with the time t in seconds according to x = 6.1t2 - 6.1t and y = 6.5t2 - t3/ 2.5. Determine the magnitudes of the velocity v and acceleration a and the angles which these vectors make with the x-axis when t = 3.5 s.Answers: When t = 3.5 s,
The particle P starts from rest at point A at time t = 0 and changes its speed thereafter at a constant rate of 1.9g as it follows the horizontal path shown. Determine the magnitude and direction of its total acceleration at times t = 0.69 s and t= 1.32 s. State your directions relative to the x-axis shown (CCW positive) and choose the angle with the smallest magnitude. B - 2.53 m- 3.62 m Answers: Att = 0.69 s, a = i 18.639 m/s?, i Att = 1.32 s, a= m/s?, e-
The position of a particle that is moving along a horizontal x-axis is given by x = (t3 - 3t2-5) m (x is positive to the right). Calculate the total distance traveled by the particle for the time interval from t = 0s to t = 4 s. O 12 m O 16 m 18 m 24 m O
2/3 The velocity of a particle which moves along the s-axis is given by v = 2 + 5t32, where t is in seconds and v is in meters per second. Evaluate the displacement s, velocity v, and acceleration a when t = 4 s. The parti- cle is at the origin s = 0 when t = 0. 214 The velocity of a particle along the s-axis is given by
A particle moving in the x-y plane has a velocity at time t = 5.25 s given by (3.9i+ 6.0j) m/s, and at t = 5.34s its velocity has become (4.05i + 6.13j) m/s. Calculate the magnitude a of its average acceleration during the 0.09-s interval and the angle it makes with the x- axis. Answers: a= m/s² 0= 2.205 i 139.1

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