4. A particle starts from an initial position with coordinates To = 8 + 5ſm, at...
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
A particle starts from the origin at t Owith a velocity of 6.2j and moves in the xy plane with a constant acceleration of (5.2i +2.6j) m/s2. At the instant the particle's x coordinate is 29 m, what are (a) its y coordinate and (b) its speed? (a) Number Unitsm (b) Number Units m/s
A particle starts at time at the position The velocity of the particle is written in the polar basis associated with its current position, and is: Matlab/Mathematica input: x0 = 13 y0 = -12 What is the position of at ? A particle P starts at time t=0 s at the position x = 13 m y = –12 m. The velocity ✓ of the particle is written in the polar basis associated with its current position, and is: ū...
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
A particle inoves along the x-axis. It's position as a function of time is given by z (t)t+22- The following questions refer to that situation. Only consider times t greater than or equal to zero fno negative values of t. Note application of the derivative is finding the maximo and minima of functions. O 1m 0 2m D Question 7 1 pts For times t between t- 0 and t 3 s, what is the minimum value of x attained...
. A particle has an initial position vector r=0 and an initial velocity v0=3i +2j(where distance is measured in meters and velocity in meters per second).The particle moves with a constant acceleration a=i -4j(measured in m/s2).At what time does the particle reach a maximum y coordinate? What is the position vector of the particle at that time?
the velocity of a particle is given by v=[16t^2i+4t^3j +(5t+2)k]m/s, where t is in seconds. If the particle is at the origin when t=0, determine the magnitude of the particle's acceleration when t=2s. What is the x,y,z coordinate position of the particle at this instant.
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 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.