4.5 Constant Acceleration


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1 4.5 Consan Acceleraion v() v() = v 0 + a a() a a() = a v 0 Area = a (a) (b) Figure 4.8 Consan acceleraion: (a) velociy, (b) acceleraion When he x componen of he velociy is a linear funcion (Figure 4.8(a)), he average acceleraion, Δv / Δ, is a consan and hence is equal o he insananeous acceleraion (Figure 4.8(b)). Le s consider a body undergoing consan acceleraion for a ime inerval [0, ], where Δ =. Denoe he x componen of he velociy a ime = 0 by v 0 v( = 0). Therefore he x componen of he acceleraion is given by Δv v() v a() = = 0. (4.4.) Δ Thus he x componen of he velociy is a linear funcion of ime given by 4.5. Velociy: Area Under he Acceleraion vs. Time Graph v() = v 0 + a. (4.4.) In Figure 4.8(b), he area under he acceleraion vs. ime graph, for he ime inerval Δ = 0 =, is Area(a(), ) = a. (4.4.3) From Eq. (4.4.), he area is he change in he x componen of he velociy for he inerval [0, ]: Area(a(),) = a = v() v 0 = Δv. (4.4.4) 4.5. Displacemen: Area Under he Velociy vs. Time Graph In Figure 4.9 shows a graph of he x componen of he velociy vs. ime for he case of consan acceleraion (Eq. (4.4.)).
2 v() v() = v0 + a A = (v() v0 ) v0 A = v0 O Figure 4.9 Graph of velociy as a funcion of ime for a consan. The region under he velociy vs. ime curve is a rapezoid, formed from a recangle wih area A = v0, and a riangle wih area A = (/ )(v() v0 ). The oal area of he rapezoid is given by Area(v(),) = A + A = v0 + (v() v0 ). (4.4.5) Subsiuing for he velociy (Eq. (4.4.)) yields Area(v(),) = v0 + a. (4.4.6) Recall ha from Example 4. (seing b = a and Δ = ), vave = v0 + a = Δx /, (4.4.7) herefore Eq. (4.4.6) can be rewrien as Area(v(),) = (v0 + a) = vave = Δx (4.4.8) The displacemen is equal o he area under he graph of he x componen of he velociy vs. ime. The posiion as a funcion of ime can now be found by rewriing Equaion (4.4.8) as x() = x0 + v0 + a. (4.4.9) Figure 4.0 shows a graph of his equaion. Noice ha a = 0 he slope is nonzero, corresponding o he iniial velociy componen v0.
3 x() ) x0 O slope = v0 Figure 4.0 Graph of posiion vs. ime for consan acceleraion. Example 4.4 Acceleraing Car A car, saring a res a = 0, acceleraes in a sraigh line for 00 m wih an unknown consan acceleraion. I reaches a speed of 0 m s and hen coninues a his speed for anoher 0 s. (a) Wrie down he equaions for posiion and velociy of he car as a funcion of ime. (b) How long was he car acceleraing? (c) Wha was he magniude of he acceleraion? (d) Plo speed vs. ime, acceleraion vs. ime, and posiion vs. ime for he enire moion. (e) Wha was he average velociy for he enire rip? Soluions: (a) For he acceleraion a, he posiion x() and velociy v() as a funcion of ime for a car saring from res are x() = (/ ) a vx () = a. (4.4.0) b) Denoe he ime inerval during which he car acceleraed by. We know ha he posiion x( ) = 00 m and v( ) = 0 m s. Noe ha we can eliminae he acceleraion a beween he Equaions (4.4.0) o obain x() = ( / ) v(). (4.4.) We can solve his equaion for ime as a funcion of he disance and he final speed giving x() =. (4.4.) v() We can now subsiue our known values for he posiion x( ) = 00 m and v( ) = 0 m s and solve for he ime inerval ha he car has acceleraed 3
4 = x( ) v( ) = 00 m = 0s. 0 m s (4.4.3) c) We can subsiue ino eiher of he expressions in Equaion (4.4.0); he second is slighly easier o use, v( ) 0 m s (4.4.4) a= = =.0 m s. 0s d) The x componen of acceleraion vs. ime, x componen of he velociy vs. ime, and he posiion vs. ime are piecewise funcions given by m s ; 0 < 0 s, a() = 0; 0 s < < 0 s  ( m s ); 0 < 0 s, v() =  0 m s ; 0 s 0 s (/ )( m s ) ; 0 < 0 s. x() =  00 m +(0 m s )( 0 s); 0 s 0 s The graphs of he x componen of acceleraion vs. ime, x componen of he velociy vs. ime, and he posiion vs. ime are shown in Figure 4.. (e) Afer acceleraing, he car ravels for an addiional en seconds a consan speed and during his inerval he car ravels an addiional disance Δx = v( ) 0s = 00 m (noe ha his is wice he disance raveled during he 0s of acceleraion), so he oal disance raveled is 300 m and he oal ime is 0s, for an average velociy of vave = a() 300 m =5m s. 0s (4.4.5) x() v() 00 m 0 m s m s 0 s 0 s 0 s 0 s 0 s 0 s Figure 4. Graphs of he xcomponens of acceleraion, velociy and posiion as piecewise funcions 4
5 Example 4.5 Caching a Bus A he insan a raffic ligh urns green, a car sars from res wih a given consan acceleraion, 3.0 m s . Jus as he ligh urns green, a bus, raveling wih a given consan velociy,.6 0 m s , passes he car. The car speeds up and passes he bus some ime laer. How far down he road has he car raveled, when he car passes he bus? Soluion: There are wo moving objecs, bus and he car. Each objec undergoes one sage of onedimensional moion. We are given he acceleraion of he car, he velociy of he bus, and infer ha he posiion of he car and he bus are equal when he bus jus passes he car. Figure 4. shows a qualiaive skech of he posiion of he car and bus as a funcion of ime. x bus x () car x () 0 a Figure 4. Posiion vs. ime of he car and bus Choose a coordinae sysem wih he origin a he raffic ligh and he posiive x  direcion such ha car and bus are ravelling in he posiive x direcion. Se ime = 0 as he insan he car and bus pass each oher a he origin when he ligh urns green. Figure 4.3 shows he posiion of he car and bus a ime. x () x () 0 + x Figure 4.3 Coordinae sysem for car and bus Le x () denoe he posiion funcion of he car, and x () he posiion funcion for he bus. The iniial posiion and iniial velociy of he car are boh zero, x,0 = 0 and v,0 = 0, 5
6 and he acceleraion of he car is nonzero a 0. Therefore he posiion and velociy funcions of he car are given by x () = a, v () = a. The iniial posiion of he bus is zero, x,0 = 0, he iniial velociy of he bus is nonzero, v,0 0, and he acceleraion of he bus is zero, a = 0. Therefore he velociy is consan, v () = v,0, and he posiion funcion for he bus is given by x () = v,0. Le = a correspond o he ime ha he car passes he bus. Then a ha insan, he posiion funcions of he bus and car are equal, x ( a ) = x ( a ). We can use his condiion o solve for : a v,0 ()(.6 0 m s  ) (/ )a a = v,0 a a = =  =. 0 s. (3.0m s ) Therefore he posiion of he car a a is a v,0 ()(.6 0 x m s  ( ) = a = = ) a a  ) =.7 0 m. a (3.0 m s 6
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