A particle moves along the x axis. Its position is given by the equation x = 1.5 + 2.9t − 3.6t2 with x in meters and t in seconds. (a) Determine its position when it changes direction. m (b) Determine its velocity when it returns to the position it had at t = 0? (Indicate the direction of the velocity with the sign of your answer.) m/s
Given that position is given by the equation
x = 15 + 2.9t - 3.6t^2
Velocity of particle will be given by:
V = dx/dt
V = d[15 + 2.9t - 3.6t^2]/dt
V = 0 + 2.9*1 - 2*3.6*t
V = 2.9 - 7.2*t
Part A.
When velocity will be zero, after that time particle will change it's direction, So
V = 0
2.9 - 7.2*t = 0
t = 2.9/7.2 = 0.403 sec
So at this time position of particle will be
x = 1.5 + 2.9*0.403 - 3.6*0.403^2
x = 2.08 m
Part B.
At t = 0, position of particle is
x = 1.5 + 2.9*0 - 3.6*0^2
x = 1.5 m
Particle will be at this same position when
1.5 = 1.5 + 2.9*t - 3.6*t^2
2.9*t - 3.6*t^2 = 0
t = 2.9/3.6 = 0.805 sec
at t = 0.805 sec, velocity of this particle will be
V = 2.9 - 7.2*0.805
V = -2.90 m/sec
Please Upvote.
A particle moves along the x axis. Its position is given by the equation x =...
A particle moves along the x axis. Its position is given by the equation x = 2.1 + 2.7t − 3.5t2 with x in meters and t in seconds. (a) Determine its position when it changes direction. m (b) Determine its velocity when it returns to the position it had at t = 0? (Indicate the direction of the velocity with the sign of your answer.) m/s
A particle moves along the x axis. Its position is given by the equation x = 1.5+ 2.5t -3.8t2 with x in meters and t in seconds. (a) Determine its position when it changes direction (b) Determine its velocity when it returns to the position it had at t =0? (Indicate the direction of the velocity with the sign of your answer.)
A particle moves along the x axis. Its position is given by the equation x = 1.8 + 2.5t − 3.8t2 with x in meters and t in seconds. (a) Determine its position when it changes direction. (b) Determine its velocity when it returns to the position it had at t = 0? (Indicate the direction of the velocity with the sign of your answer.) I have a) as 2.21m, i am having issues solving part (b).
Average and Instantaneous Velocity A particle moves along the x axis. Its position varies with time acording to the expression x =-4t + 2t2, where x is in meters and t is in seconds. The position-time graph for this motion is shown in the figure. Notice that the particle moves in the negative x direction for the first second of motion, is momentarily at rest at the moment t = 1 s, and moves in the positive x direction at times...
A particle moves along the x axis according to the equation x = 1.93 + 2.90t − 1.00t2, where x is in meters and t is in seconds. (a) Find the position of the particle at t = 3.10 s. m (b) Find its velocity at t = 3.10 s. m/s (c) Find its acceleration at t = 3.10 s. m/s2
A particle moves along the x axis according to the equation x = 2.06 + 2.95t - 1.0062, where x is in meters and t is in seconds. (a) Find the position of the particle at t = 2.80 s. m (b) Find its velocity at t = 2.80 s. m/s (c) Find its acceleration at t = 2.80 s. m/s2 Submit Answer
A particle moves along the x axis according to the equation x = 1.93 + 2.99t-1.00p, where x is in meters and t is in seconds. (a) Find the position of the particle at t2.60 s. (b) Find its velocity at t -2.60 s m/s (c) Find its acceleration at t-2.60 s m/s2
Two particles move along an x axis. The position of particle 1 is given by x = 6.00t2 + 4.00t + 5.00 (in meters and seconds); the acceleration of particle 2 is given by a = -9.00t (in meters per seconds squared and seconds) and, at t = 0, its velocity is 21.0 m/s. When the velocities of the particles match, what is their velocity?
Two particles move along an x axis. The position of particle 1 is given by x = 8.00t2 + 3.00t + 6.00 (in meters and seconds); the acceleration of particle 2 is given by a = -10.0t (in meters per seconds squared and seconds) and, at t = 0, its velocity is 22.0 m/s. When the velocities of the particles match, what is their velocity?
a particle moves along the x axis. its position as a function of time is given by x = 6.8 t + 8.5 t^2 , where t is in seconds and x is in meters. what is the acceleration as a function of time?