The velocity of a particle moving in the x-y plane is given by (6.07i+ 4.97j) m/s at time t = 3.87 s. Its average acceleration during the next 0.014 s is (7.5i + 5.0j) 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.
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- The velocity of a particle moving in the x-y plane is given by (7.481 +6.64)) m/s at time t = 6.53 s. Its average acceleration during the next 0.012 s is (5.8i + 4.3)) m/s². Determine the velocity v of the particle at t = 6.542 s and the angle between the average- acceleration vector and the velocity vector at t = 6.542 s. Answers: v=i i i+ j) m/sThe velocity of a particle moving in thex-y plane is given by (4.03i + 7.36j) m/s at time t = 3.64 s. Its average acceleration during the next 0.020 s is (2.2i + 2.6j) m/s?. Determine the velocity v of the particle at t = 3.660 s and the angle e between the average- acceleration vector and the velocity vector at t = 3.660 s. Answers: v= ( i i+ i j) m/s e =The equation r(t) = ( sin t)i + ( cos t)j + (t) k is the position of a particle in space at time t. Find the particle's velocity and acceleration vectors. π Then write the particle's velocity at t= as a product of its speed and direction. The velocity vector is v(t) = (i+j+ k.
- The acceleration of a particle moving only on a horizontal xy plane is given by á = 3t i+4t j, where á is in meters per second-squared and t is in seconds. At t = 0, the position vector is 7 = (20 m) i+(40 m) j and the velocity vector v = (5 m/s) i+(2 m/s) j. Find expressions for the (a) x and (b) y component (in meters) of the position of the particle at time t. (c) What is the angle between its velocity and the positive direction of the x axis at time t? (а) х %3D 5t+20) i m Edit (b) y = Edit (c) e = Edit Click if you would like to Show Work for this question: Open Show WorkThe position r of a particle moving in an xy plane is given by ř seconds. In unit-vector notation, calculate (a) 7, (b) V , and (c) a for t = 3.00 s. (d) What is the angle between the positive direction of the x axis and a line tangent to the particle's path at t = 3.00 s? Give your answer in the range of (-180°; 180°). (4.00r3 – 1.00t)î + (5.00 – 1.00r4)j with 7 in meters and t in (a) Number i i Units (b) Number ît i Units i (c) Number i i Units (d) Number i UnitsAt time t = 0, the position vector of a particle moving in the x-y plane is r = 4.62i m. By time t = 0.011 s, its position vector has become (4.84i +0.54j) m. Determine the magnitude Vay of its average velocity during this interval and the angle 0 made by the average velocity with the positive x-axis. Answers: Vav 0= i i m/s
- The velocity of a particle moving in the x-y plane is given by (7.15i +4.12j) m/s at time t = 5.37 s. Its average acceleration during the next 0.022 s is (3.9i+ 6.9j) m/s2. Determine the velocity v of the particle at t = 5.392 s and the angle between the average- acceleration vector and the velocity vector at t = 5.392 s. Answers: v = ( i 0= i it i j) m/sThe velocity vector of a particle is given by: V = Vje-)[sin(wt)i + cos(wt)j] where, Vo = 12.1 m/s; T = 2.5 s; w = 5.2 rad/s Calculate the magnitude of the acceleration (in m/s2) at t = 7.4 sThe velocity of a particle moving in the x-y plane is given by (6.98i + 6.68j) m/s at time t = 7.60 s. Its average acceleration during the next 0.014 s is (4.1i + 5.4j) m/s2. Determine the velocity v of the particle at t = 7.614 s and the angle θ between the average-acceleration vector and the velocity vector at t = 7.614 s.Answers:v = ( _____ i + _____ j) m/sθ = _____ °
- The acceleration of a particle moving only on a horizontal xy plane is given by d = 6tî + 7tj, where d is in meters per second-squared and tis in seconds. At t= 0, the position vector 7 = (15.0m)î + (35.0m)ĵ locates the paticle, which then has the velocity vector V = (6.40m/s)î + (3.30m/s)ĵ. At t = 5.60 s, what are (a) its position vector in unit-vector notation and %3D (b) the angle between its direction of travel and the positive direction of the x axis? (a) Number i j Units i (b) Number i UnitsThe velocity of a particle moving in the x-y plane is given by (6.12i + 3.24j) m/s at time t = 3.65 s. Its average acceleration during the next 0.02 s is (4.0i + 6.0j) m/s². Determine the velocity v of the particle at t = 3.67 s and the angle between the average-acceleration vector and the velocity vector at t = 3.67 s. Answers: V = ( i 8= i it i j) m/sAn object has an initial velocity of 29.0 m/s at 95.0° and an acceleration of 1.90 m/s2 at 200.0°. Assume that all angles are measured with respect to the positive x-axis. (a) Write the initial velocity vector and the acceleration vector in unit vector notation. (b) If the object maintains this acceleration for 12.0 seconds, determine the average velocity vector over the time interval. Express your answer in your unit vector notation.