v(t) = t³ - 10t² + 29t-20, 1 st≤ 6 (a) Find the displacement. x (b) Find the total distance that the particle travels over the given interval. X
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- At t = 0 the coordinates of a particle are x = 0, y = 2.10 m. At a time 1.05 s later, its coordinates are x = 2.94 m, y = 3.80 m. Find the components of the particle's average velocity during this time interval. vx = ? vy = ?The position coordinate of a particle which is confined to move along a straight line is given by "x = 2t3 −24t + 6", where x is measured in meters from a convenient origin and t is in seconds. Determine (a) the time required for the particle to reach a velocity of 72 m/s from its initial condition at t = 0 s, (b) the acceleration of the particle when v = 30 m/s, and (c) the net displacement of the particle during the interval from t = 1 s to t = 4 s.The velocity function, in feet per second, is given for a particle moving along a straight line, where t is the time in seconds. Find (a) the displacement and (b) the total distance that the particle travels over the given interval. v(t) = 5t − 7, 0 ≤ t ≤ 3
- 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 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?The velocity function v(t) (in meters per second) is given for a particle moving along a line. v(t) = 3t-10, 0 less than or equal to t less than or equal to 4 (a) Find the displacement d1 traveled by the particle during the time interval given above.d1 = m(b) Find the total distance d2 traveled by the particle during the time interval given above.d2 = m
- The velocity function, in feet per second, is given for a particle moving along a straight line. v(t) = t2 − t − 90, 1 ≤ t ≤ 15 (a) Find the displacement.(b) Find the total distance that the particle travels over the given interval.An animal has a coordinates of (1.1 m, 3.4 m) at time t1 = 0 and when t2= 3.0 s, it is located at (5.3 m, -0.5 m). For this time interval, find (a) the components of the average velocity, and (b) the magnitude and direction of the average velocity.The velocity function of a particle moving along a horizontal line is given by v(t) = ln 2 * 23t , where t ≥ 0 is in seconds. The particle is initially 2/3 units to the left of the origin. Find the position of the particle when the acceleration is equal to 6(ln 2)2 .
- An objects position in the x-direction as a function of time is given by the expression: x(t)=5t^2 + 2t where are quantities have proper SI units. What's the objects average velocity in the x-direction between the times t=1.15s and t=2.07s. (Just enter the number rounded to 3 significant figures and assume it has proper SI units)The velocity function, in feet per second, is given for a particle moving along a straight line, where t is the time in seconds. Find (a) the displacement and (b) the total distance that the particle travels over the given interval. v(t) = t3 − 10t2 + 27t − 18, 1 ≤ t ≤ 7A 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.