If we know thex acce1eration ax as a function of time and we know the initial

If we know thex acce1eration ax as a function of time

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If we know thex-acce1eration ax as a function of time and we know the initial velocity VOx, we can use Eq. (2.17) to find the x-velocity Vx at any time; in other words, we can find Vx as a function of time. Once we know this function, and given the initial position xo, we can use Eq. (2.18) to find the position x at any time. Motion with changing acceleration Sally is driving along a straight highway in her classic 1965 Mus- tang. At tinte t = 0, when Sally is moving at 10 m/s in the positive x-direction, she passes a signpost at x = 50 m. Her x-acceleration is a function of tinte: ax = 2.0m/sz - (O.lOm/s')t (a) Find her x-velocity and position as functions of tinte. (b) When is her x-velocity greatest? (c) What is the maximwn x-velocity? (d) Where is the car when it reaches the maximwn x-velocity? IDENTIFY: The x-acceleration is a function of tinte, so we cannot use the constant-acceleration formulas of Section 2.4. SET UP: We use Eqs. (2.17) and (2.18) to find the x-velocity and position as functions of time. Once we have those functions, we'll be able to answer a variety of questions about the motion. EXECUTE: (a) At t = 0, Sally's position is Xo = 50 m and her x-velocity is v", = 10 m/s. Since we are given the x-acceleration ax as a function of time, we firstuseEq. (2.17) to find the x-velocity Vx as a function of tinte t. The integral of t" is Jt" dt = " ! 1 t"+1 for n * -1,so Vx = lOm/s + I'[2.0m/s' - (O.lOm/s')tldt o 1 = lOm/s + (2.0m/s')t - 2(0.lOm/s')t' Then we use Eq. (2.18) to find x as a function of t: x = 50m + fJ lOm/s + (2.0m/s')t - ~(O.lOm/s')t'l dt 1 I = 50m + (lOm/s)t + 2(2.0m/s')t' - 6(0.lOm/s')t' Figure 2.29 shows graphs of a .. v .. and x as functions of time. Note that for any tinte t, the slope of the vx-t graph equals the value of ax and the slope of the x-t graph equals the value of V x . (b) The maximum value ofv x occurs when the x-velocity stops increasing and begins to decrease. At this instant, du,/dt = ax = O. Setting the expression for ax equal to zero, we obtsin 0= 2.0m/s' - (O.lOm/s')t 2.0m/s' t=--- =20s O.lOm/s' (c) We find the maximum x-velocity by substituting t = 20 s (when x-velucity is maximwn) into the equation for Vx from part (a): 1 v""",.x = lOm/s + (2.0m/s')(20s) - 2(0.1Om/s')(20s)' = 30m/s
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2.29 The position, velocity, and acceleration of the car in Example 2.9 as functions of time. Can you show that if this motion continues, the car will stop at t = 44.5 s? ax (mfs2) 2.0 L ...... x-accelemtion is ~ ~~ti~~ before t = 20 s. 1.0 ~ -:::O+---l.5-..JI0--IL5-2O-'--~-'-S-..J30- t(s) -1.0 x-acce1emtion is .... .... ............ negative after t = 20 s. t=208. 0 5 10 15 t(s) 20 25 30 x(m) 800 600 400 25 30 t(8) 2.6 *Velocity and Position by Integration 59 (d) The maximum value of Vx occurs at time t = 20 s. To obtain the position of the car at that time, we substitnte t = 20 s into the expression for x from part (a): I x = 50m + (1Om/s)(20 s) + 2:(2.0m/ S 2)(20 S)2 - ~(O.IO m/ s 3)(20 S)3 = 517m EVALUATE: Figure 2.29 helps us interpret our results. The top graph in this figure shows that ax is positive between t = 0 and t = 20 s and negative after that. It is zero at t = 20 s, the time at which Vx is maximum (the high point in the middle graph). The car speeds up until t = 20 s (because Vx and ax have the same sign) and slows down after t = 20 s (because Vx and ax have opposite signs).
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