A particle's position with respect to time as it moves along a coordinate axis is given by the function p(t)=t3+2t−2. What is the particle's acceleration at time t=−1?
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A particle's position with respect to time as it moves along a coordinate axis is given by the function p(t)=t3+2t−2. What is the particle's acceleration at time t=−1?
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- The position of a particle moving along the x axis varies in time according to the expression x = 3t2, where x is in meters and t is in seconds. Evaluate its position (a) at t = 3.00 s and (b) at 3.00 s + t. (c) Evaluate the limit of x/t as t approaches zero to find the velocity at t = 3.00 s.A particle moves along the x axis. Its position is given by the equation x = 2 + 3t 4t2, with x in meters and t in seconds. Determine (a) its position when it changes direction and (b) its velocity when it returns to the position it had at t = 0.At time t=0s, an object is moving to the right with a velocity v that can be modeled by the equation v=(4.2 m/s)-(1.4 m/s2)t. At what time, if any, does the object change its direction of motion?
- A particle performs a one-dimensional motion with the position given by the time equation x(t) = 1,3t4 - 2,0t3, where x is given in meters and t is given in seconds. At time t=0, the particle starts its motion at the origin x=0. At what instant (in seconds) does the particle reverse its direction of motion?During the time interval 0 < t < 1, the velocity of an object as a function of time is v = 5 + 3t^2 + 2t^3. Knowing that the object follows a rectilinear trajectory, What is the correct expression for the acceleration?A particle is moving horizontally with constant acceleration. The position (x, in meters) as a function of time (t, in seconds) is given by x= 0.798 t^2 + 0.3552 t + 0.0355. Based on this information, the acceleration of the particle is what?
- 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.A particle moves along the x axis according to the equation x=2.00 + 3.00t - 1.00t², where x is in meters and t is in seconds. At t=3.00s, find (a) the position of the particle, (b) its velocity, and (c) its acceleration.A particle moves along the x-axis according to the equation x = 2.00 +3.00t -1.00t^2, where x is in meters and t is in seconds. At t = 3.00s , find the acceleration of the particle.
- If an object initially at rest is given a velocity of: vx =-1.93 m/s and vy = 2.51 m/s. What is the magnitude and direction its displacement after a time of 3.50 seconds?On a spacecraft two engines fire for a time of 567 s. One gives the craft an acceleration in the x direction of ax = 4.80 m/s2, while the other produces an acceleration in the y direction of ay = 7.52 m/s2. At the end of the firing period, the craft has velocity components of vx = 3751 m/s and vy = 4834 m/s. Calculate the magnitude of the initial velocity.A particle moves along the positive x-axis and its position as a function of time is x(t)=(19.9 m/s)t − (1.7 m/s^2)t^2. What is the average velocity in the time interval between 2.3s and 3.7s in m/s?