During a time interval of two seconds, a particle changed its velocity from V = (2i -3j) m/s to V = (10i +5 j) m/s. The magnitude of the particle's average acceleration (in m/s?) during this time interval is: %3D %3D A. 0.45 B. 5.66 C. 2.24 D. 9.8 E. 12.3
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- A particle’s velocity along the x-axis is described byv(t) = A t + B t2,where t is in seconds, v is in meters per second, A = 1.06 m/s2, and B = -0.51 m/s3. What is the distance traveled, in meters, by the particle between times t0 = 1.0 s and t1 = 3.0 s?A jeep travels a distance d=22.1m in the positive x direction in a time t1=20.2s, at which point the jeep brakes, coming to rest in t2=7.38s. 1. What was the jeep's instantaneous velocity in the horizontal direction, in meters per second, when it began braking? 2. Using the result from question 1, what was the jeep's horizontal component of acceleration, in meters per squared second, during the braking period?An object is moving along the x-axis. At t = 0 it is at x = 0. Its x-component of velocity Vx as a function of time is given by: Vx(t) = at - Bt3, where a = 6.8 m/s2 and B = 4.0 m/s4 I. At t = 1.8 s, what is the x-component of the velocity of the object? (Express your answer with the appropriate units.) II. At t = 1.8 s, what is the x-component of the acceleration of the object? (Express your answer with the appropriate units.)
- An unidentified flying object (UFO) is observed to travel a total distance of 19000 m, starting and ending at rest, over a duration of 4.23 s. Assuming the UFO accelerated at a constant rate to the midpoint of its journey and then decelerated at a constant rate the rest of the way, what was the magnitude of its acceleration? Express your answer in g s , where 1 g = 9.81 m/s^2.A ball is thrown into the air by a baby alien on a planet in thesystem of Alpha Centauri with a velocity of 49 ft/s. Its heightin feet after t seconds is given by y = 49t −25t2.a.) Find the average velocity for the time period beginningwhen t0 = 1 second and lasting for the given time.t = .01 sec:t = .005 sec:t = .002 sec :t = .001 sec: b.) Estimate the instanteneous velocity when t = 1.On a spacecraft two engines fire for a time of 532 s. One gives the craft an acceleration in the x direction of ax = 4.92 m/s2, while the other produces an acceleration in the y direction of ay = 7.16 m/s2. At the end of the firing period, the craft has velocity components of vx = 3743 m/s and vy = 4808 m/s. Calculate the magnitude of the initial velocity.
- At t=0 an object starts from the origin and moves with a constant acceleration of a→=(−3.2i−2.7j)m/s2 on a horizontal surface. If the initial velocity of the object is vo→=(14.1i−18.3j)m/s, determine the magnitude of velocity at t=2.1s. Express your answer in units of m/sm/s using zero decimal places.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 = 3.00 s, find (a) the position of the particle, (b) its velocity, and (c) its acceleration.A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 146 feet per second from an initial height of 89 feet off the ground, then the height of the projectile, hh, in feet, tt seconds after it's shot is given by the equation: h=−16t2+146t+89h=-16t2+146t+89 Find the two points in time when the object is 126 feet above the ground (to one decimal places). Answer: The object is 126 feet off the ground at the following times: (Enter your two answers separated by a comma. Round to ONE decimal places.)
- A small object moves along the xx-axis with acceleration ax(t)ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At tt = 0 the object is at xx = -14.0 mm and has velocity v0xv0xv_0x = 6.00 m/sm/s.. A).What is the xx-coordinate of the object when tt = 10.0 ss?An object's velocity as a function of time in one dimension is given by the expression; v(t) = 2.71t + 7.88 where are constants have proper SI Units. What is the object's velocity at t = 4.19 s?A ball kicked from ground level at an initial velocity of 68 m/s and an angle θ with ground reaches a horizontal distance of 295 meters. What is the size of angle θ and what is time of flight of the ball?a. θ = 17.223⁰, t = 8.44 sb. θ = 19.372⁰, t = 4.6 sc. θ = 18.263⁰, t = 9.3 sd. θ = 16.273⁰, t = 7.3 s