INSTANTANEOUS VELOCITY I claim ha ha if acceleraion is consan, hen he elociy is a linear funcion of ime and he posiion a quadraic funcion of ime. We wan o inesigae hose claims, and a he same ime, work ou wha we mean by insananeous elociy. Assume ha he relaion beween disance and ime is gien by x() 5. Suppose we wan o find he (insananeous) elociy a sec. We sar wih a graph. x (m) 80 60 posiion x s. ime 0 3 (sec) Fig. Le s find he aerage elociy. In general, x( ) x( ) a Le sec and 3 sec. Then he aerage elociy is jus he slope of he line (dashed line in drawing) connecing hese wo poins: 5 5 5 9 5 a 0 m/s 3 To find he insananeous elociy, le s make his ime incremen smaller and smaller. Firs, le sec and sec. ()
a 5 5 5 5 5 m/s Now le s ry.5 sec. Then a 5(.5) 5.5 m/s. And so on..5 Exercise: Show ha for.,.0,... sec, a 0.5, 0.05, m/s. The aerage elociy seems o be geing closer and closer o 0 m/s, as we ges closer and closer o. Will his paern hold? Le s inesigae he quesion algebraically. For our funcion, he aerage elociy will be 5 5 a. Now, le s define a small ime incremen Δ ha we can make as small as we wan. We can now wrie +Δ, and our expression for aerage elociy becomes Wih a lile algebra, his equaion becomes a a +Δ 5 5 +Δ 5 5 5 + 0 Δ + 5Δ 5 +Δ Δ +Δ 0Δ+ 5Δ Δ 0 + 5Δ Exercise: Show ha he appropriae choice of we found aboe. If we now le ime : Δ gies he same alues for he aerage elociy Δ shrink o zero, he aerage elociy becomes he insananeous elociy a he 0 where for conenience I hae dropped he subscrip. In mahemaical language, we hae aken a limi, and in he process hae aken a deriaie he slope of he line angen o he cure a any arbirary ime. The insananeous elociy is hus he deriaie of he posiion. We hae also shown ha if he posiion is a quadraic funcion of ime, hen he elociy is linear in he ime as claimed. Exercise: Can you go on o show ha he acceleraion is consan in his example?
We hae also raised a whole hos of mahemaical quesions: Hae I inaderenly diided by zero aboe, for example? More generally, does his limi always exis? These quesions and ohers we may safely leae o your mahemaics courses! A non-calculus approach o insananeous elociy Galileo go hese same resuls, decades before Newon and Leibniz inened he calculus. Here is (more or less) how he did i. We firs noice ha if he acceleraion is consan, a graph of elociy s. ime will gie a sraigh line. To undersand his poin, recall ha he aerage acceleraion is jus he slope of a graph of elociy s. ime; so if he graph of elociy s. ime is a sraigh line, he slope he aerage acceleraion always has he same alue. So now we know ha if acceleraion is consan, he elociy is a linear funcion of ime. If we can show ha posiion is a quadraic, we will be done! We will need he mean speed heorem his heorem was known mahemaically in he Middle Ages, and Galileo applied i o moion. I saes he following: (m/s) (m/s) elociy s. ime elociy s. ime 0 a 0 0 0 3 (sec) Fig. (a) 3 Fig. (b) (sec) Mean Speed Theorem: If acceleraion is consan, he disance raeled in a ime ineral is he same as he disance he objec would rael a he consan mean speed he aerage of he iniial and final elociies for he same ime ineral. (Noe: YF alks abou his heorem, wihou giing i a name, on page 8 see especially Equaion -0.) To see why his heorem is rue, consider he graphs of elociy s. ime shown aboe: In he firs graph (Fig (a)), he disance raeled by an objec moing a he mean speed is he mean speed muliplied by he ime ineral: where mean iniial final Δ x Δ mean ( + )/. This resul should be eiden from he figure. In oher words, he aerage elociy defined in Eq. () is equal o he mean speed he aerage of he iniial and final elociies ONLY if he acceleraion is consan. Noe from Fig. ((a) ha his mean speed is also he insananeous speed a he middle of he ime ineral. 3
Noice ha his disance, Δ x, is jus he area of he recangle shown in he graph. Now consider he second graph (Fig. (b). The areas of he green and gray riangles in he second graph are equal. Inuiiely, he disance raeled by he uniformly acceleraed objec is likewise he area under he line (gray area in Fig. (b). Therefore, he shaded areas in Fig. (a) and Fig. (b) are he same: The uniformly acceleraed objec raels he same disance ha i would if i were moing a he (consan) mean elociy. Or in oher words, he mean elociy is jus he aerage elociy as defined in Eq. (). We hae jus proen he mean speed heorem. The res is sraighforward: If we call and, hen he disance iniial final raeled is Δ x aδ ( + ) Δ since, as we hae jus shown, a ( + ). As usual, Δ. Bu from our definiion of aerage acceleraion, a. Afer a line or wo of algebra, his equaion becomes and herefore + a( ) + aδ Δ x ( + ) Δ ( + [ + aδ] ) Δ Δ + a ( Δ ) Tha resul is wha we expec. I s a lile simpler if we le Δ (ha is, we le our clock sar a 0 and le. In ha case we hae Δ x + a. Thus, wihou using calculus, we hae shown ha if acceleraion is consan, elociy is a linear funcion of ime, and displacemen a quadraic funcion of ime! Compare his resul o YF Eq. (.)
Finally, consider Fig. 3, shown below. The acceleraion is no longer consan, so he (m/s) elociy s. ime 0 a 0 3 Fig. 3 elociy is no longer linear. Noice ha he mean elociy is he same as before: he aerage of he elociy a s and 3 s. Bu he areas of he wo riangular segmens are no longer equal: The mean speed heorem does no work for his case i works only when he acceleraion is consan! (sec) 5