particle starts from the origin at t = 0 with a velocity of 6.0[pic] m/s and moves in the xy plane with a constant acceleration of (-2.0[pic] + 4.0[pic]) m/s2. At the instant the particle achieves its maximum positive x coordinate‚ how far is it from the origin? [pic] 2 At t = 0‚ a particle leaves the origin with a velocity of 5.0 m/s in the positive y direction. Its acceleration is given by [pic] = (3.0[pic] - 2.0[pic]) m/s2. At the instant the particle reaches its maximum y coordinate how far
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being applied to the object should appear on that object’s free-body diagram. We should include a downward normal force‚ applied to the elevator by you. Yes‚ mg is numerically equal to this normal force in this case. When the system has an acceleration‚ however‚ these forces are no longer equal. The system has a constant velocity directed up When the system of you and the elevator is moving up with a constant velocity‚ what do we need to change on the freebody diagrams? 1. An extra
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Biomechanical Analysis of Skilled Movement Sprint start INTRODUCTION Sprinting is a dynamic sport‚ the aim is to produce as much power and momentum in order to maximise the speed and velocity in which a sprinter covers a distance of 100 meters. There are many factors that will determine the speed and as a result the time obtained in a race. The start‚ or the sprint start as it is known‚ is a vital part of the 100 meter sprint. It determines how fast the sprinter will be going at the start of the
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inclined plane. To measure the instantaneous velocity and to determine the acceleration of the cart from the slope of the velocity-time graph. Theoretical Background A cart moving down a smooth incline speeds up. This is a simple case of a uniformly accelerated motion in one dimension. The rate of change of velocity is constant or uniform. The rate of change of velocity is called acceleration. To determine the acceleration‚ one needs to measure the velocity at two different points along the incline
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stationary with a flat line across the 0.7 line. Acceleration graph begins sloping negatively once the force of hand is applied. After 2.6 seconds the cart is pushed towards the sensor until it reaches 0.2 meters. At this point the power of fan becomes greater than the power of the hand and the cart changes direction. Net force equals Fhand. All three graphs show this movement with a negative sloping and then a positive sloping in Acceleration halfway through Region B which in turn makes velocity
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field experiment about speed‚ a scientist created the chart above. The chart shows distance and time measurements for a racing A. 0 m/s B. 96 m/s C. 192 m/s D. 384 m/s 2. What is the racing car’s acceleration? A. 0 m/s2 B. 96 m/s2 C. 192 m/s2 D. 384 m/s2 3. Theodore plots the data in the chart above. He plots distance on the y-axis and time on the x-axis. How will the movement of the racing
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subtopics of kinematics and dynamics. Kinematics is concerned with the aspects of motion that exclude the forces that cause motion. In a manner of speaking‚ kinematics is focussed on the development of definitions: position‚ displacement‚ velocity‚ acceleration and on the relationships that exist between them. Dynamics widens the study of motion to include the concepts of force and energy. Definitions Position Kinematics begins with the idea of position. Suppose that we photograph an object moving
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Subjective Srivastava’s Srivastava’s MOTION IN ONE DIMENSION 1. A particle is projected vertically upwards with a velocity 20 m s1 from the top of the tower of height 100 m. Determine the time it takes to reach the ground. Answer: t 2 1 6 sec. A particle is dropped from the top of the tower of height h. At the same instant another particle is projected vertically upwards from the bottom of the tower with such a velocity that it will be able to just reach the top of the tower. When and at what
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with a constant acceleration‚ reaching a height of 63 m in 8 s. * Speed at height 63 - 2x63/8 = answer What is the acceleration= answer/8 * If you were to drop a rock from a tall building‚ assuming that it had not yet hit the ground‚ and neglecting air resistance‚ after 5.5 s; * How fast? 5.5x10 * How far? Half gravity = 5x5.5^2 * A 1200 kg racing car accelerates from rest at a constant rate and covers a distance of 400 m in 10 s. What is the car’s acceleration? (in m/s2) 400/10
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The average acceleration‚ a = ᐃv/ ᐃt‚ is in the same direction as ᐃv‚ that is‚ toward the center of the circle. As the object moves around the circle‚ the direction of the acceleration vector changes‚ but its length remains the same.One should take note of the fact that the acceleration vector of an object in uniform circular motion always points in toward the center of the circle. Due to this fact‚ the acceleration of such an object is called center- seeking or centripetal acceleration. Remember
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