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- At time t = 0, a particle is located at the point (4,8,6). It travels in a straight line to the point (1,7,2), has speed 2 at (4,8,6) and constant acceleration - 3i -j- 4k. Find an equation for the position vector r(t) of the particle at time t.Find the vector equation of the line that passes through the point (1, 5) and is parallel to the line r = (3 + 7t)i + (1 − 3t)j.A particle starts at the origin with initial velocity 7 - j+ 3k. Its acceleration is å(t) = 6t i+ 12t° j - 6t k. Find the velocity and position vectors for this particle.
- The position vector r(t) = 4ti + 4tj − 2tk describes the path of an object moving in space . (a) Find the velocity vector, speed, and acceleration vector of the object. (b) Evaluate the velocity vector and acceleration vector of the object at the given value of t=3.An object has the velocity vector function i(t) = (5, 4e“, 2t + 6) and initial position 7 (0) = (– 2, 1, – 3) A) Find the vector equation for the object's position. F(t) > B) Find the vector equation for the object's acceleration. a(t) = >4 Suppose the position vector of a particle in space at time t is given by: r = (12t,-312,-8t/2) (a) Find the particle's velocity and acceleration vectors. (b) Find the particle's speed and direction of motion.
- find the vector equation for the line through the point A = (5, -2, 1) at t = 0. and B = (-3, 1, -2) at t = 1.Find a vector equation for the line through the point P = (-3, 5, -4) and parallel to the vector v = (5, -1, 1). Assume r(0) = -3+5-4k and that v is the velocity vector of the line. k r(t) =Find the coordinates of a point P on the line and a vector v parallel to the line (x + 3)/5 = y/8 = (z − 3)/6
- At time t = 0, a particle is located at the point (8,4,7). It travels in a straight line to the point (3,9,6), has speed 3 at (8,4,7) and constant acceleration negative 5i + 5 j - k. Find an equation for the position vector r(t) of the particle at time t.Given the vector function 7 (t) = (- 4t, – 5t", 2t? + 2) Find the velocity and acceleration vectors at t = 2. v(2) = a(2) =Find the cartesian equation of the straight line tangent to the plane curve given by the vector equation r(t) = (t- 6) i + + 3t - 2) į. at the point P(-5,4) on the curve. Express your answer in the form y = ax + b , where a and b are integers. equarion of the tangent line is given by y = 9x + 49