The vector position of a particle varies in time according to the expression F(t)=(1.41 t + 1.72)i +(0.224t2 + 1.11)J where r is in meters and t is in seconds. Find the speed of the particle in m/s at 2.58 s.
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- At time t = 0 a particle at the origin of an xyz-coordinate system has a velocity vector of Vo =i+ 12j - k. The acceleration function of the particle is a(t) = 16ti+j+ (cos 2t)k. Find the speed of the particle at time t = 1. Round your answer to two decimal places. Speed =the coordinates of an object moving in an xy plane vary with the time according to the eqations x=-7.85 sin w(omega)t, and y=4-7.85 cosw(omega)t. W(omega) is constant , x,y - in meters. And t in seconds. Write expressions for the possition,velocity and acceleration vectors of the object at any time t>0.The vector position of a particle varies in time according to the expression 7(t)=(1.76 t + 1.56)7 +(0.452t2 + 1.34)Ï where r is in meters and t is in seconds. Find the speed of the particle in m/s at 2.19 s.
- A space shuttle’s coordinates as functions of time are given by x(t) = 7.0 t^3 and y(t) = 4.0t^2 - 2.1t , where x and y are in meters and t is in seconds. Write a vector expression for the ball’s velocity as a function of time.The velocity i of a particle moving in the xy plane is given by 7 = ( 5.50 - 5.00 ji + 9.00 3. with V in meters per second and t (> 0) in seconds. At I- 1.40 s and in unit-vector notation, what are (a) the x component and (b) the y component of the Eceleratiun? (c) Whien(if uver) is the acceluration zero? (d) At wlat pusitive time does the speed equal 10.0 mh? (a) Numher 7 Unit (b) Number 1 Unit {c) Number Unit (d) Number UnitVectors u = −10i + 3j and v = −7i − 9j. What is u − v? a −17i − 6j b 17i + 6j c 3i − 12j d −3i + 12j
- 5. A particle moves counterclockwise on a circular path of 400 ft radius. It starts from a fixed point which is horizontally to the right of the center of the path and moves so that s=1012 + 20t where s is the arc distance in feet and t is the time in sec. Compute the x and y components of acceleration at the end of 3 sec.At time t = 0 a particle at the origin of an xyz-coordinate system has a velocity vector of vo = i+ 5j – k. The acceleration function of the particle is a(t) = 32r²i + j+ (cos 21)k. Find the speed of the particle at time t = 1. Round your answer to two decimal places.√√3 =(√³1) 1+ (1-161²) The equation r(t) = vectors at time t = 0. j is the position of a particle in space at time t. Find the angle between the velocity and acceleration The angle between the velocity and acceleration vectors at time t= 0 is (Type an exact answer, using as needed.) radians.
- A velocity vector has x-component vx = 6.3 m/s and y-component vy = - 2.5 m/s. What is the total magnitude of the velocity? O 3.8 m/s 6.7 m/s 5.8 m/s 8.8 m/sThe position of an object as a function of time is given by r = (3.2t + 1.5t2)i + (1.7t -2.0t2)jm, where t is time in seconds. What is the magnitude of the acceleration, and what is the direction of the acceleration (given in terms of theta, clockwise from the x axis).A space shuttle’s coordinates as functions of time are given by x(t) = 7.0 t^3 and y(t) = 4.0t^2 - 2.1t , where x and y are in meters and t is in seconds. write a vector expression for the ball's position as a function of time.