I need accurate answers to this , with detailed step s Examples 7A particle moves in a straight line with a constant acceleration of -4 ft/s2 for 6 s, zeroacceleration for the next 4 s, and a constant acceleration of +4 ft/s2 for the next 4 s. Knowingthat the particle starts from the origin and that its velocity is -8 ft/s during the zero-accelerationtime interval, (a) construct the v - t and x - t curves for 0 sts 14 s, (b) determine the positionand the velocity of the particle and the total distance traveled when t = 14 s.a (ft/s²)40-461014t(s)

Step 1: Formulate the velocity model Since the acceleration is constant, we can conveniently use the…

Answered: Examples 7 A particle moves in 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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I need solutions to this calculation questions please
Q No 1: The position of a particle moving along the x-axis is given by x=A+B t3, where A=9.75 cm and B=1.50cm/s3.consider the time interval t=2 to t=3s and calculate (a).the average velocity; (b).the instantaneous velocity at t=2s;(c). the instantaneous velocity at t=3s;(d). the instantaneous velocity at t=2.5s;(e). the instantaneous velocity when the particle is midway between its positions at t=2s and t=3s.
A turtle crawls along a straight line, which we’llcall the x-axis with the positive direction to the right. The equationfor the turtle’s position as a function of time is x 1t2 = 50.0 cm +(2.00 cm/s)t - (0.0625 cm/s2)t2. Sketch graphs of x versus t, vx versus t, and ax versus t, for the time interval t = 0 to t = 40 s.
The velocity V of a particle moving in the xy plane is given by V = (5.5t - 4.2t2) i + 8.7 j, with V in meters per second and t (> 0) in seconds. At t = 1.9 s and in unit- vector notation, what are (a) the x component and (b) the y component of the acceleration? (c) When (if ever) is the acceleration zero? (d) At what positive time does the speed equal 10 m/s?
Please help me..
A police car is traveling at a velocity of 20.0 m/s due north, when a car zooms by at a constant velocity of 47.0 m/s due north. After a reaction time 0.800 s the policeman begins to pursue the speeder with an acceleration of 3.00 m/s^2. Including the reaction time, how long does it take for the police car to catch up with the speeder? Number i Units Vos A VOS North AVP AVOP AVOP Police car catches up with speeder ↑ ap Police car begins to accelerate - Speeder passes police car
Initially (at t=0 s) a weather balloon is at x = 0 and at time t (seconds) its horizontal velocity component is 0,2t m/s. The equation of its path is y = 0,2x² m. (a) Calculate the magnitude and direction of its velocity at time t = 10 s. (b) Calculate the radius of curvature at x = 10 m as well as the magnitude of the normal directed acceleration component an there. (c) Assume the acceleration a = 2,4 m/s² and the magnitude of the normal component an 0,4 m/s². Calculate the tangential component at.
i need and detailed answer to a-d thankyou^^
The figure below (not to scale) shows an acceleration-time graph of a body. a (m/s^2) 0 7 t(s) B 27 If the initial velocity was zero, a(0) = a0 = 10.7 m/s^2, a(7) = a7 = -2.2 m/s^2 and the particle's acceleration returns to zero at t = 10s, what will the body's velocity be at t = 10 s? Express your answer to three significant figures in units of m/s (but do not write the units with your answer).
I need the answer quickly
1) Most of us rode bikes a kid. Did you ever accidentally apply the front brake while riding down a hill and flip over your handlebars? Let's analyze the level of deceleration needed to cause this to happen (at this instant the normal force of the rear tire is zero). Assume the rider is moving down a hill at a constant velocity and att = 0, the rider squeezes the brakes and begins to decelerate. This action results in a force that points in the bi direction and passes through the contact points between the tires and the pavement. Be sure to include your FBD, IRD, and transformation array. b2 30 in. FB 13 in. 23 in.
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