The position of a particle is given by F = (7.7t+7)i - 3tj+5k, where F in (m) and t in (s). The magnitude of the velocity (in m/s) at t= 19.6 s is:
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- The x-coordinate of a particle in curvilinear motion is given by x = 3.9t3 - 1.6t where x is in feet and tis in seconds. The y-component of acceleration in feet per second squared is given by a, = 5.2t. If the particle has y- components y = 0 and vy = 3.7 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. %3D Answers: V = i ft/sec a = i ft/sec?At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of i = (3.00 i − 2.00 j) m/s and is at the origin. At t = 4.00 s, the particle's velocity is = (9.10 i + 6.50 j) m/s. (Round your coefficients to two decimal places.) (a) Find the acceleration of the particle at any time t. (Use the following as necessary: t.)b) Find its x and y coordinates at any time t. (Use the following as necessary: t.)The x-coordinate of a particle in curvilinear motion is given by x = 4.7t3 - 5.5t wherexis in feet and t is in seconds. The y-component of acceleration in feet per second squared is given by a, = 3.3t. If the particle has y-components y = 0 and vy = 4.4 ft/sec when t= 0, find the magnitudes of the velocity v and acceleration a when t = 3.8 sec. Sketch the path for the first 3.8 seconds of motion, and show the velocity and acceleration vectors for t = 3.8 sec. Answers: V = i ft/sec a = ft/sec?
- A particle which moves with curvilinear motion has coordinates in millimeters which vary with the time t in seconds according to x = 1.9t²-3.1t and y-6.1t2-t3/1.2. Determine the magnitudes of the velocity v and acceleration a and the angles which these vectors make with the x-axis when t - 4.3 s. Answers: When t = 4.3 s, V= a- i mm/s, ex- mm/s², 0,- iThe x-coordinate of a particle in curvilinear motion is given by x= 2.1t³ -2.4t where x is in feet and t is in seconds. The y-component of acceleration in feet per second squared is given by ay = 2.0t. If the particle has y-components y = 0 and vy= 2.2 ft/sec when t = 0, find the magnitudes of the velocity v and acceleration a when t = 2.7 sec. Sketch the path for the first 2.7 seconds of motion, and show the velocity and acceleration vectors for t = 2.7 sec. Answers: V = i a= i ft/sec ft/sec²A particle is initially at rest and begins to accelerate to the right as 13t-t^2 m/s. How far does the particle move in 3 seconds? the instructor told me I had to integrate, the velocity is not constant.
- The position of a particle is described byr = (t3+ 4t –4) m and θ = (t3/2) rad, where t is inseconds. Determine the magnitudes of the particle’svelocity and acceleration at the instant t = 2 s.A particle has a position in space given by r = 2t^3 i + 4t^2 j - 3t k where r is in m and t in seconds. Find the magnitude of the instantaneous velocity at t = 3.0 s. 24 m/s 75 m/s 54 m/s 59 m/sA particle confined to the xy-plane is in uniform circular motion around the origin. The x- and y-coordinates of the particle vary with time as follows: x(t) = r sin(cot) y(t) = r cos(cot). where >= 2.60 m and c = 2.15 s-¹. What are the x- and y-components of the particle's acceleration at an instant when x = 1.00 m and y = 2.40 m. Give your answers by entering numbers into the empty boxes below. ax || || X x² ms ms-2.
- IT.0.0a Given an acceleration vector, initial velocity (uo.Vo), and initial position {xo.yo}, find the velocity and position vectors for t2 0. a(t) = (cos t,5 sin t. (Uo Vo) = (0,5) , (Xo.Yo) = (2.0) What is the velc | vector? v(t) =Initially, a bicycle is at the position 7 = (130 m)i + (210 m)j. After 120 s, the position of the particle is 7 = (110 m)i + (240 m)j . What is the average velocity vector of the bicycle in that time || tinterval? O Tav = (0.17 m/s)i + (-0.25 m/s)j O Jay = (0.25 m/s)i + (-0.17 m/s) O Tay = (-0.25 m/s)î + (0.17 m/s) O Jay = (-0.17 m/s)i + (0.25 m/s)jPLEASE ANSWER QUICKLT AND WILL RATE HIGHLY *15