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The three-dimensional motion of a particle is defined by the position
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Vector Mechanics For Engineers
- Your answer is partially correct. The velocity of a particle moving in the x-y plane is given by (7.70i + 4.91j) m/s at time t = 7.38 s. Its average acceleration during the next 0.029 s is (3.3i + 2.6j) m/s?. Determine the velocity v of the particle at t = 7.409 s and the angle 8 between the average- acceleration vector and the velocity vector at t = 7.409 s. Answers: v = ( 7.7957 i+ 4.9754 j) m/s 5.73arrow_forwardAt time t = 0, the position vector of a particle moving in the x-y plane is r = 5.28i m. By time t = 0.022 s, its position vector has become (5.39i+ 0.48j) m. Determine the magnitude Vay of its average velocity during this interval and the angle 8 made by the average velocity with the positive x-axis. Answers: Vav 0 = = 20 i m/sarrow_forwardclassical dynamicsarrow_forward
- Subject Dynamicsarrow_forwardAt time t = 0, the position vector of a particle moving in the x-y plane is r = 4.78i m. By time t = 0.019 s, its position vector has become (4.96i + 0.52j) m. Determine the magnitude Vay of its average velocity during this interval and the angle e made by the average velocity with the positive x- axis. Answers: Vav Ꮎ = = i i m/sarrow_forwardAt time t=0, the position vector of a particle moving in the x-y plane is r = 4.48i m. By time t = 0.030 s, its position vector has become (4.711 +0.47)) m. Determine the magnitude vay of its average velocity during this interval and the angle 9 made by the average velocity with the positive x-axis. Answers: i m/s 4arrow_forward
- A girl is playing fetch and catch with her dog. If the ball leaves the girl’s hand with the initialvelocity as shown in the diagram, and the dog is running along the positive x-direction with a velocity of 6m/s, determine the relative velocity of the ball with respect to the dog at the instant the ball is thrown.Report your answer in Cartesian vectors.arrow_forwardAn automobile P is traveling along a circular track of radius R=958.4 m. At position "A" on the track, the automobile has a speed of UA = 10.3 m/s. At this position, the driver of the automobile applies the brakes causing the speed of the automobile to change with distance s traveled along the track according to the following equation: U(S) = VA COS(0.001s) m/s (cos is in radians), where s is given in meters. Determine the magnitude of the acceleration for the driver when the automobile reaches position "B" on the track where "B" is a quarter of the distance around the track from position "A". R B O circular trackarrow_forwardA revolving slotted arm OA moves a bearing P in a fixed curve shape: r(θ) = Cθ. r = radial distance to O θ = angle the arm OA makes in the x-direction C = known constant Arm OA begins from rest when θ = π/4 and rotates counterclockwise with constant angular acceleration d2θ/dt2 = α. Part 1: Calculate v(t) (the velocity vector) of the P bearing as a function of t. Show result with respect to C, α, and the unit vectors ur (r-direction) and uθ (θ-direction). Part 2: Calculate a(t) (acceleration vector) of the P bearing as a function of t. Show result with respect to C, α, and the unit vectors ur (r-direction) and uθ (θ-direction). Part 3: Calculate magnitude v of the velocity and the magnitude a of the acceleration of P when the angle of the slotted arm is θ = 3π/4.arrow_forward
- A motorcycle cage has a curved path in the x-z plane as shown. The cage path has different curved profile, and between point o and A the path is best described as z = f(x) = ax², where a = 0.09 and b = 1.23. The motorcyclist at point M is moving at a velocity of 12.6 m/s which is increasing at a rate of 1.9 m/s?. If the horizontal position of the motorcyclist at point M is xM = 11.9 m, determine (i) the direction of the motorcyclisť's velocity. (ii) the radius of the curvature of the cage's path at point M, (iii) the normal acceleration at point M, (iv) the magnitude and direction of the motorcyclist's acceleration at point M. A (z = axarrow_forwardA particle moves in a circle with a diameter of 120 cm. If its linear velocity is given by v = 4.5t where v is in mps and t is in s, find the radial, tangential and the resultant acceleration of the particle.arrow_forward5 (a)The x-coordinate of a particle in curvilinear motion is given by x = 3.7t3 - 4.5t where x is in feet and t is in seconds. The y-component of acceleration in feet per second squared is given by ay = 1.7t. If the particle has y-components y = 0 and vy = 3.4 ft/sec when t = 0, find the magnitudes of the velocity v and acceleration a when t = 5.9 sec. Sketch the path for the first 5.9 seconds of motion, and show the velocity and acceleration vectors for t = 5.9 sec.arrow_forward
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