23A truck covers 40.0 m in 8.50 s while smoothly slowing down to a final speed of 2.80 m/s.(a)Find its original speed(b)Find its acceleration

(a)vxf = vxi + atxf = xi + vxit + at2vxf2 = vxi2 + 2a(xf xi)xf = xi + (vxf + vxi)t / 2

xi = 0xf = 40.0 mvxi = ?vxf = 2.80 m/sa = ?t = 8.5 s

2.80 = vxi + a(8.5)Equation has vxi, a (Solve second)40.0 = 0+vxi(8.5) + a(8.5)2Equation has vxi, a2.802 = vxi2+2a(400)Equation has vxi, a40.0 = 0 + (2.80 + vxi)(8.5) / 2Equation has vxi (Solve first)

vxi = 40 x 2/8.5 2.80 = 6.61 m/s

(b)2.80 = 6.61 + a(8.5)a = (2.80 6.61)/8.5 = -0.45 m/s2

25.An object moving with uniform acceleration has a velocity of 12.0 cm/s in the positive x direction when its x coordinate is 3.00 cm. If its x coordinate 2.00 s later is -5.00 cm, what is its acceleration?

vxf = vxi + atxf = xi + vxit + at2vxf2 = vxi2 + 2a(xf xi)xf = xi + (vxf + vxi)t / 2

xi = 3.00 cmxf = -5.00 cmvxi = 12..0 cm/svxf = ?a = ?t = 2.00 s

vxf = 3.00 + a(2.00)Equation has vxi, a-5.00 = 3.00 + 12.0(2.00) + a(2.00)2Equation has avxf2 = 12.02+2a(-5.003.00)Equation has vxi, a

-5.00 = 3.00 + 12(2.00) + a(2.00)2-8.00 = 24.0 + a(4)a = -32.0 / 2.00 = 16.0 cm/s227A particle moves along the x axis. Its position is x = 2 + 3t - 4t2, with x in meters and t in seconds. Determine(a)its position when it changes direction(b)its velocity when it returns to the position it had at t = 0

(a)x = 2 + 3t 4t2 xf = xi + vxit + at2Compare the two equations to findxi = 2.00 mvxi = 3.00 m/sa = -8.00 m/s2

v = dx/dt = 3 8tDirection changes when v = 00 = 3 8tt = -3/-8 = 0.375 s

x = 2 + (3 x .375) 4(.375)2x = 2 + 1.125 0.563 = 2.56 m

(b)xf = xi + vxit + at2xf = xi0 = vxit + at20 = vxi + att = -2vxi /a

t = 2vxi /-8.00 = -2(3)/-8 = 3/4 sv = 3 8t = 3 8(3/4) = -3.00 m/s

24.

35. Un camin en un camino recto parte del reposo y acelera a razn de 2 m/s2 hasta alcanzar una velocidad de 20 m/s. Despus, el camin viaja durante 20 s con velocidad constante hasta que aplica los frenos para detener el camin de manera uniforme en 5 s ms. (a) Cunto tiempo permanece el camin en movimiento?. (b) Cul es la velocidad media del camin en el movimiento descrito?.

Rta: a) 35 s; b) 15,71 m/s

x1 = 1/2at1 ^ 2v = t1 => t1 = v / a = 20/2 = 10 sx1 = 0.5 * 2 * 100 = 100 mx2 = vt2t2 = 20 sx2 = 20 * 20 = 400 mx3 = vt3-1/2at3 ^ 2t3 = 5 sa = v/t3 = 20/5 = 4 m / s ^ 2x3 = 20 * 5-0,5 * 4 * 25 = 50 mt = t1 + t2 + t3 = 10 +20 +5 = 35 s [contestar]x = x1 + x2 + x3 = 100 +400 +40 = 550 mvelocidad arage esv = x / t = 550/35 = 15,7 m / s [Respuesta b]

Distancia = velocidad promedio * TiempoVelocidad media = * (vi + vf)de distancia = * (vi + vf) t *Para la primera distancia, t = (20-0) 2 = 10 segundosprimera distancia = * ( 0 + 20) * 10 = 100 m2 de distancia = 20 m / s * 20 = 400 mEn los ltimos 5 segundos, la velocidad del camin disminuy de 20 m / s a 0 m / s.tercera distancia = * (20 + 0) * 5 = 50 mLongitud total = 550 metrosTiempo total = 35 segundosVelocidad media = 550 35La velocidad media es de aproximadamente 15,7 m / s

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