Un automóvil circula a una velocidad de 60 km/h durante 1 hora y 15 minutos, después se para durante 5 minutos y luego regresa hacia el punto de partida a una velocidad de 10 m/s durante 45 minutos. Halla: a) La posición final. b) La distancia recorrida. c) La velocidad media.

(a) 48 km

(b) 102 km

(c) 48.96 km/h.

Explanation:

Assuming that at first, the car travels from point A to point B, then the cat returns from point B to point C.

As the car traveled from point A to point B in 1 hour 15 minutes with a speed of 60 km/h.

Using distance = speed x time, so

Distance traveled from A to B is

km [ as time = 1 h 15 min = 1.25 h]

Stop time at point B is 5 minutes.

For the return journey, from point B to C:

Speed = 10 m/s

Time = 45 minutes = 45 x 60 seconds = 2700 seconds

Distance traveled from point B to C is

10 x 2700 = 27000 meters

as 1000 meters = 1 km, so

27 km.

(a) The final position of the car from the starting point A is the distance between points A and C.

75 -27=48 km.

Hence, the car is 48 km away from point A.

(b) Total distance traveled,

75+27=102 km.

(c) Total time taken

t= 1.25 hour + 5 minutes + 45 minutes

= 5/4 hour + 5/60 hour + 45/60 hour [ as 1 hour = 60 min]

=25/ 12 hours.

Total distance traveled, d= 102 km [ from part (b)]

Average speed = (Total distance traveled)/(Total time taken)

=102/ (25/12)= 48.96 km/h.

Hence, the average speed of the car is 48.96 km/h.

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

a 2,000 calorie based diet.