A particle moves along the x axis. Its position varies with time according to th

Question

A particle moves along the x axis. Its position varies with time according to the expression x = -4t + 2t2, where x is in meters and t is in seconds. The position-time graph for this motion is shown in the figure. Notice that the particle moves in the negative x direction for the first second of motion, is momentarily at rest at the moment t = 1 s, and moves in the positive x direction at times t > 1 s.

(C) Find the instantaneous velocity of the particle at t = 2.5 s. Measure the slope of the green line at t = 2.5 s (point ) in the figure: vx= m/s Notice that this instantaneous velocity is on the same order of magnitude as our previous results, that is, a few meters per second. Is that what you would have expected?Suppose a particle moves along the x axis. Its x coordinate varies with time according to the expression x = 16 + 3.9t + 2.4t2, where x is in meters and t is in seconds.

(a)Determine the displacement ?x of the particle in the time interval t = 9 s to t = 20 s. ?x= m Your response differs from the correct answer by more than 10%. Double check your calculations.

(b) Calculate the average velocity in the time interval t = 9 s to t = 20 s. v = m/s

(c) Find the instantaneous velocity of the particle at t = 18.4 s. v = m/s

Explanation / Answer

I'm answering the 2nd question: Suppose a particle moves along the x axis. Its x coordinate varies with time according to the expression x = 16 + 3.9t + 2.4t2, where x is in meters and t is in seconds. a) plug in t = 9 , then X = 245.5m plug in t = 20 to your equation, then X = 1054m b) Use the equation x = x0 + v0t + 1/2 at^2 to find that a = 2.4 * 2 = 4.8 and v0 = 3.9 Use the equation v = v0 + at to find the velocity at 9 seconds and at 20 seconds V at 9 seconds = 3.9 + 4.8 * 9 = 47.1 m/s V at 20 seconds = 99.9 m/s average velocity = (v0 + v)/2 = 73.5 m/s c)Use the equation v = v0 + at to find that v = 3.9 + 18.4 * 4.8 = 92.22 m/s

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