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A particle moves along a straight line such that its acceleration is a = ( 4t2 – 2) m/s2 , where t is in seconds. When t = 0, the particle is located 2 m to the left of the origin, and when t = 2 s, it is 20 m to the left of the origin. Determine the position of the particle when t = 4 s.
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- The velocity of a particle which moves along the s axis is given by v = 7 - 4t + 7t3/2, where t is in seconds and v is in meters per second. Evaluate the position s, velocity v, and acceleration a when t = 4 s. The particle is at the position so = 4 m when t = 0. Answers: S= V = a = i i i m m/s m/s²arrow_forwardThe velocity of a car traveling along a straight line is given by v = (4 · t² – 7 · t) · ft where t is in S seconds. Initially, s 2 ft when t = 0. %3D a. Determine the position s of the car when t = 7 s. b. Determine the total distance travelled during the time interval t = 0 s to t = 7 s. c. Determine the acceleration a at t = 2 s .arrow_forward3. The motion of the particle along a straight line is governed by the relation a = t³ – 2t2 + 7 where a is the acceleration in m/s? andt is the time in seconds. At time t velocity of the particle is v m/s and the displacement is d m. Calculate the displacement, velocity and acceleration at time t = 2 seconds. * A second, the 0.5 А second V = ..m/s d 25 marrow_forward
- 60 ft/s Q1: Car A starts from rest at straight road with a constant acceleration of 4 ft/s? until it reaches a speed of 70 ft/s. Afterwards it maintains this speed. Also, when t = 0, car B located 6000 ft down the road is traveling towards A at a constant speed of 60 ft/s. Determine the distance traveled by car A when they pass t = 0 and travels along a 6000 ft each other.arrow_forwardThe acceleration of a particle is defined as a = 9 – 3t2. The particle starts at t 0 with v = 0 and x = 5 m. Determine: 1. The time when the velocity is again zero. 2. The position and velocity when t = 4 s Select one: A. t = 3; x(4) = 13 m and v(4) = -28 m/s %3D B. t = 1; x(4) = 13 m and v(4) = -28 m/s %3D !i! C. t = 0; x(4) = 13 m and v(4) = -28 m/s %3D %3D D. t = 3; x(4) = 13 m and v(4) = 28 m/s %3Darrow_forwardText:The particle travels along the path defined by the parabola y=0.5x². If the component of velocity along the x-axis is v=5t m/s, where "t" is in seconds, determine the particle's distance from the origin and the magnitude of its acceleration when t=1s. When t=0, x,=0, yo=0. y=0.5xarrow_forward
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