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satela [25.4K]
3 years ago
15

You drive 6.0 km at 50 km/h and then another 6.0 km at 90 km/h. Your average speed over the 12 km drive will be ____

Physics
1 answer:
Elis [28]3 years ago
5 0

Answer:

The average speed is less than 70 km/h.

(B) is correct option.

Explanation:

Given that,

distance = 6.0 km

Speed = 50 km/h

Speed = 90 km/h

We need to calculate the time in 6.0 km distance

Using formula of time

t = \dfrac{d}{v}

Put the value in to the formula

t=\dfrac{6.0}{50}

t=0.12\ hr

We need to calculate the time in another distance

Using formula of time

t = \dfrac{d}{v}

Put the value in to the formula

t=\dfrac{6.0}{90}

t=0.067\ hr

We need to calculate the average speed

Using formula of average speed

v=\dfrac{D}{T}

Where, D = total distance

T = total time

Put the value into the formula

v=\dfrac{12}{0.12+0.067}

v=64.17\ km/hr

Hence, The average speed is less than 70 km/h.

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12. A las 10 de la mañana Elena sale a 100 Km/h de una ciudad A con dirección a Madrid. A la misma hora sale Javier desde otra c
Klio2033 [76]

Answer:

a) t = 3.3 [h]

b ) Hora = 13:18 o 1:18 [pm]

c) x = 327.79 [km/h] (Elena)

x = 196.6 [km] (Javier)

Explanation:

Para poder solucionar este problema debemos hacer un planteamiento inicial de ubicacion de las ciudades, este planteamiento nos ayudara a entender el problema de una manera mas facil.

Tenemos las ciudades A & B y la ciudad de Madrid que esta a una distancia x con respecto de B, (ver esquema adjunto).

De manera logica debemos deducir que la Ciudad A debe estar mas lejos de Madrid que la ciudad B de la misma Madrid, ya que en caso contrario Javier nunca alcanzara a Elena, ya que Elena va mas rapido que Javier.

a) Ahora debemos de utilizar la siguiene ecuacion de la cinematica, cuando los cuerpos se mueven a velocidad constante.

x=x_{o}+v*t

Donde:

x -xo = Distancia entre el punto inicial y punto final.

v = velocidad [m/s]

t = tiempo [s]

Debemos convertir las velocidades de kilometros por hora a metros por segundo.

100 [\frac{km}{h} ]*1[\frac{h}{3600s}]*1000[\frac{m}{1km} ] = 27.77[m/s]\\60[\frac{km}{h} ]*1[\frac{h}{3600s}]*1000[\frac{m}{1km} ] = 16.66[m/s]

Seguidamente formulamos una ecuacion por cada movimiento, luego debemos igualar estas ecuaciones en funcion de la variable x que sera el punto donde se encuentren ambas personas.

<u>Para Elena</u>

(132000+x) = 27.77*t\\x = 27.77*t - 132000

<u>Para Javier</u>

<u />x - xo = 16.66*t\\xo = 0\\x = 16.66*t<u />

Igualamos las variables x de ambas ecuaciones.

16.66*t = 27.77*t -132000\\27.77*t - 16.66*t = 132000\\11.11*t = 132000\\t = 11880.83 [s] = 3.3 [h]

b) La hora facilmente se puede encontrar sumando el tiempo con las 10:00am

hora = 10 + 3 = 13 [hrs]

La parte decimal debe convertirse a tiempo.

0.3 [hr]*60[\frac{min}{1hr} ]= 18 min

Hora = 13:18 o 1:18 [pm]

c) Para encontrar estas distancias utilizamos el tiempo encontrado en el item a.

<u>Para Elena</u>

x = v*t\\x = 27.77*11800.83 = 327791.6 [m] = 327.79 [km]\\

<u>Para Javier</u>

<u />x = v*t\\x = 16.66*11800.83 = 196601.8 [m] = 196.6 [km]<u />

6 0
3 years ago
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