A particle starts from the origin at t Owith a velocity of 6.2j and moves in...
A particle starts from the origin at t = 0 with a velocity of (8.3j hat) m/s and moves in the xy plane with a constant acceleration of (3.2i hat + 1j hat) m/s2 . At the instant the x coordinate of the particle is 29 m, A. what is the value of its y coordinate and B. its speed
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/s2. At the instant the x coordinate of the particle is 15 m, what is the speed of the particle? 10 m/s 16 m/s 12 m/s 14 m/s 26 m/s
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
4. A particle starts from an initial position with coordinates To = 8 + 5ſm, at time t= 0, with a velocity of V. = 3i-8 m/s. The particle moves in the r-y plane with a constant acceleration, à = -21 - m/s. (a) At the instant the y-coordinate of the particle's position is -10 m, find the x- coordinate of its position. (b) Calculate the x- and y-components of the particle's position when the particle reaches its turning point...
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 s, a particle starts from the origin of a coordinate system and moves in the xy plane with a velocity ~v = (7.9ˆi − 3.2ˆj) m/s. Determine the x position of the particle at t = 2.0 s.
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 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 starts from the origin at t = 0 and moves along the positive x axis. A graph of the velocity of the particle as a function of the time is shown in the figure; the v-axis scale is set by vs = 6.0 m/s. (a) what is the coordinate of the particle at t = 5.0 s? (b) what is the velocity of the particle at t = 5.0 s? (c) what is the acceleration of the particle...
please i need help asap Problem 1 The acceleration of a particle moving only on a horizontal xy plane is given by a=3ti+4tj, where a is in meters per seconds squared and t is in seconds, at t=0, the position vector r=(20.0m)i+(40.0m)j locates the particles, which then has the velocity vector v=(5.00m/s)i+(2.00m's)j. at t=4.00s, what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the x axis?...