To find velocity we need to differentiate x(t) with respect to t.
After that check the sign of v(t) = x'(t) on the interval 0 <
t <
.
If the sign is positive then velocity is increasing and vice versa.
2. Each of four particles moves along an x axis. Their coordinates (in meters) as functions...
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 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 = 1.5 + 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
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 with velocity given by v(t) = 32 – 1 for time t > 0. If the particle is at position x = 5 at time t = 0, what is the position of the particle at timet 1?
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. 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).
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