expression x= 3t³ – 2t +5, where x is in meters and t is in seconds. | a) Determine the displacement of the particle in the time intervals t=1s to t=3s. b) Calculate the average velocity in the time intervals t=1s to t=3s.
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- The position of a particle moving along the x axis is given by x=3t2-15t-10m, where t is in s, Determine the following a. average velocity during the time interval t=2.5 s to t=4.5s b. instantaneous velocity at 3s c. average acceleration during the time interval t=2.5 s to t=4.5s d. instantaneous acceleration at 3s e. is the particle speeding up or slowing down at 4s? justify your answerA particle is undergoing a uniform rectilinear motion and the displacement as a function of time is given by r(t) = 6t4- 2t3 -12t2+3t+3 where r is in meters and t is in seconds. What is a uniform motion? Find the average velocity of the particle between t = 0 and t = 2s When is the acceleration zero?According to y(t) = 5.0t 2 − 4.0t3m, the particle’s position as it moves along the y-axis varieswith time. (a) Solve for the velocity and acceleration of the particle as functions of time, (b) thevelocity and acceleration at t = 3.0 s, (c) the time at which the position is a maximum, (d) the timeat which the velocity is zero, and (e) the maximum height.
- The velocity of a particle at any time t moving along an x-axis is given by the following equation: v = 3t^2 - 8t + 3 where v is in m/s. At t = 0, x = 0. Determine the following:(a) the particle's displacement between t = 0 and t = 4 s (b) the average velocity for the time interval t =0 to t = 4 s (c) the acceleration when t =3 s (d) the instantaneous velovity at t = 4 s (e) what is the initial direction of motion of the particle?A particles coordinates are given by x(t)=(5.0m/s)t y(t)=3.5m-(9.8 m/s²)t² Where x and y are in meters and t in seconds Calculate 1 the magnitude of the average acceleration of the particle between t=0s and t=2s. 2. Calculate the magnitude of the instantaneous velocity of the particle at time t=1s.The position of a particle moving along the x-axis varies with time according to x(t) = 5.0t2 − 4.0t3 m. Find (a) the velocity and acceleration of the particle as functions of time, (b) the velocity and acceleration at t = 2.0 s, (c) the time at which the position is a maximum, (d) the time at which the velocity is zero, and (e) the maximum position.
- The velocity function of a particle moving along a horizontal line is given by v(t) = t2-8t+15, where t >= 0 is in seconds. The particle is 5/3 units to the left of the origin at the second instant when it changes direction. 1. Determine th etime interval (s) on which the particle is both moving to the right and slowing down. 2. Find the position function of the particle.The velocity function v(t) (in meters per second) is given for a particle moving along a line. v(t) = 3t-10, 0 less than or equal to t less than or equal to 4 (a) Find the displacement d1 traveled by the particle during the time interval given above.d1 = m(b) Find the total distance d2 traveled by the particle during the time interval given above.d2 = mA particle moves from an initial position with coordinates x = -2.0 m, y = +3.0 m, to a final position with coordinates x = +5.0 m, y = +7.0 m in a time interval of 10 s. What is the magnitude of the average velocity for this displacement?
- The position of a particle moving along the x axis depends on the time according to the equation x = ct4 - bt6, where x is in meters and t in seconds. Let c and b have numerical values 2.7 m/s4 and 1.4 m/s6, respectively. From t = 0.0 s to t = 1.7 s, (a) what is the displacement of the particle? Find its velocity at times (b) 1.0 s, (c) 2.0 s, (d) 3.0 s, and (e) 4.0 s. Find its acceleration at (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s.The position coordinate of a particle which is confined to move along a straight line is given by "x = 2t3 −24t + 6", where x is measured in meters from a convenient origin and t is in seconds. Determine (a) the time required for the particle to reach a velocity of 72 m/s from its initial condition at t = 0 s, (b) the acceleration of the particle when v = 30 m/s, and (c) the net displacement of the particle during the interval from t = 1 s to t = 4 s.Using SI base units, a particle’s position as a function of time is given by x(t)=(7t3+4t2+6) m. (a) At what time will the particle’s acceleration be zero? (b) Find the speed of the particle at that time.