Question 1 A particle leaves the origin with an initial velocity v 7.94 m/s and a...
A particle leaves the origin with an initial velocity = (6.93) = 6.931 m/s and a constant acceleration (-- - 4.601 – 1.87j m/s2 When the particle reaches its maximum x coordinate, what are (a) its velocity, (b) its position vector? (a) Number Units (b) Number + Units
A particle leaves the origin with an initial velocity of v-([vO])i m/s where vO-8.25 and a constant acceleration a = (-[az])i + (-lay)j m/s2, where ax-1.89 and ay-258. When the particle reaches its maximum x coordinate, how far (in m) is it from the origin?
(8c4p 17) A particle leaves the origin with an initial velocity v 3.28, in m/s. Įt experiences a constant acceleration a =-1.00i-0.90j, in m/s. What is the velocity of the particle when it reaches its maximum x coordinate? i-component of velocity? 0.00 m/s You are correct. Your receipt no. is 166-4230 revious Tries j-component of velocity? -2.95 m/s Incorrect. Tries 5/5 Previous Tries When does it reach its maximum x coordinate? -2.952 ms Submit Answer You have entered that answer...
A particle leaves the origin with an initial velocity v Overscript right-arrow EndScripts equals left-parenthesis 8.47 i Overscript ̂ EndScripts right-parenthesis m divided by s and a constant acceleration a Overscript right-arrow EndScripts equals left-parenthesis negative 1.24 i Overscript ̂ EndScripts minus 4.99 j Overscript ̂ EndScripts right-parenthesis m divided by s Superscript 2. When the particle reaches its maximum x coordinate, what are (a) its velocity, (b) its position vector?
A particle starts from the origin at t = 0 with an initial velocity of 5.0 m/s along the positive x axis. If the acceleration is (–3.0i + 4.5j) m/s^2, determine the velocity and position of the particle at the moment it reaches its maximum x coordinate. Can someone explain why when the particle reaches it maximum x coordinate?
A particle starts from the origin at t = 0 with an initial velocity of 5.5 m/s along the positive x axis.If the acceleration is (-2.9 i^ + 4.7 j^)m/s2, determine (a)the velocity and (b)position of the particle at the moment it reaches its maximum x coordinate
At t = 0, a particle leaves the origin with a velocity of 9 0 m/s in the positive y direction and moves in the xy plane with a constant acceleration of (2.0i - 4.0j)m/s^2. At the instant the x coordinate of the particle is 15 m, what is the speed of the particle? 10m/S 16 m/s 12 m/s 14 m/s 26 m/s
A particle leaves the origin with an initial velocity v⃗ =(2.40m/s)x^, and moves with constant acceleration a⃗ =(−1.90m/s2)x^+(3.20m/s2)y^. How far does the particle move in the x direction before turning around? What is the particle's velocity at this time? Calculate the particle's position at t = 0.500 s, 1.00 s, 1.50 s, and 2.00 s. Use these results to sketch x and y positions versus time for the particle.
A particle leaves the origin with initial velocity v⃗ o=8.0i^+15j^m/s, undergoing constant acceleration a⃗ =−1.5i^+0.25j^m/s2. In what direction is it moving counterclockwise from the positive x-axis ?
Question 1 In the figure, two particles are launched from the origin of the coordinate system at time t = 0. Particle 1 of mass my - 5.20 g is shot directly along the x axis (on a frictionless floor), where it moves with a constant speed of 10.4 m/s. Particle 2 of mass m2 - 3.80 g is shot with a velocity of magnitude 17.0 m/s, at an upward angle such that it always stays directly above particle 1...