the position of a particle moving along the x-axis depends on the time according
ID: 1736752 • Letter: T
Question
the position of a particle moving along the x-axis depends on the time according to the equation x=ct^2-bt^3, where x is in meters and t in seconds. What are the units of (a) constant c and (b) constant b? Let their numerical values be 3.0 and 2.0 respectively. (c) at what time does the particle reach its max. positive x position? From t=0.0s to t=4.0s, (d) what distance does the particle move and (e) what is its displacement? Find its velocity (f) 1.0s, (g) 2.0s, (h) 3.0s, and (i) 4.0s. Find its acceleration at times (j)1.0s, (k)2.0s, (l) 3.0s, and (m) 4.0s. I need help with all the steps from c and on as i have some answers but i dont know if im using the right equations and numerical values in the right places. I am also completely lost in this chapter.Explanation / Answer
You wouldn't happen to go to CSUN, would you? The class therehas this problem in this week's assignment... . For part (c), you can find the time for the max positivedisplacement by taking dx/dt and setting itto zero: . dx/dt = 6 t - 6t2 = 0 . t ( 1 - t) = 0 t = 0 or 1 . These represent the times the particle reaches its minimum ormaximum position. To determine which, conside the velocity of theparticle at t = 0.5 sec: . v = dx/dt = 6 * 0.5 - 6 * 0.52 = 1.5 it is positive...which means between time 0 and 1 the particle is moving inthe positive direction (and at t = 1) the velocity is zero. . This means that t = 1 must bethe time the particle attains maximum positive position. . (d) this is tricky... from t = 0to 1 the velocity is positive and the particle moves apositive distance of: . x = 3 * 11 - 2*13 = 1 meter . then from time = 1to 4, the velocity is negative and the particle moves from x= 1 to... . x = 3 *42 - 2 * 43 = - 80 meter . so between t = 0 to t = 4, theparticle goes from x = 0 to 1, then from 1 to-80... a total distance of . 1 + 81 = 82meters . (e) but its displacement isonly final position - initial position = -80 - 0 = -80 meters . (d) this is tricky... from t = 0to 1 the velocity is positive and the particle moves apositive distance of: . x = 3 * 11 - 2*13 = 1 meter . then from time = 1to 4, the velocity is negative and the particle moves from x= 1 to... . x = 3 *42 - 2 * 43 = - 80 meter . so between t = 0 to t = 4, theparticle goes from x = 0 to 1, then from 1 to-80... a total distance of . 1 + 81 = 82meters . (e) but its displacement isonly final position - initial position = -80 - 0 = -80 metersRelated Questions
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