By S. B. Kizlik
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Extra info for Algebra, part 2 (tablets)
01813429400 ey - O. ez u: =dcp/sqrt (dcp . 9674163783 ey O. ez As seen from a stationary observer outside the train, the ball's speed is the sum of the component of the train's velocity iJt in the direction of it (thus, the dot product iJt· '12) plus the speed V with which the boy threw the ball. 2531907412 ex speed:=(vt. 63954584 - CHAPTER 1. VECTORS AND KINEMATICS 44 The ball's speed is about 14 m/s. Mike now summarizes the main features by making a plot. He first creates a functional operator A for producing arrows to graphically represent the various vectors in the problem.
Greg will work with the short form as illustrated in the following command line. > distance: =sqrt (sep . sep); #dot product distance := (4d 2 + 8db Vot - 8db VoT + 8dt Vo + 4b 2 Vo 2 t 2 - 8b 2 Vo 2 tT + 8 b Vo 2 t 2 + 4 b2 Vo 2 T2 - 8 b Vo 2 T t + 4 t 2 Vo 2 + 4 c2 Vo 2 t 2 - 8 c2 Vo 2 t T + 4 c2 Vo 2 T2 + 4g 2 t 2 T2 - 4g 2 tT 3 + g2 T4)(l/2) /2 Eoj's and Enaj's velocities, vi and v2, are calculated at time t > vl:=diff(rl,t); v2:=diff(r2,t); vi := - Voe x - tge z v2 := b Vo ex + c Vo ey - (t - T) 9 ez The smoke jumpers fall under the influence of Erehwon's gravity until the angle between the velocity vectors v2 and viis 90 0 • So Greg calculates the dot product between these vectors and sets the product equal to zero.
F[fr] :=mu[k]*Nj #frictional force Ffr:= J-lkN Since the toboggan is being pulled at constant speed, by Newton's first law the pull component up the slope must balance the sum of the weight component and frictional force down the slope. This condition is entered in eq2. > eq2:=P*cos(phi)=w*sin(theta)+F[fr] j #parallel to surface eq2 := Pcos(¢) = w sin(B) + J-lk N The set of simultaneous equations, eql, eq2, are analytically solved for the set of unknowns, P and N, the solution being labeled sol and the result assigned.
Algebra, part 2 (tablets) by S. B. Kizlik