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a bee flies at 20 feet per second directly to a flower bed from its hive. HThe bee stays at the flower bed for 15 minutes, then flies directly back to the hive at 12 feet per second. It is away from the hive fore 20 minutes total.
a. What equation can you use to find the distance of the flowerbed from the hive?
b. How far is the flowerbed from the hive?

Respuesta :

we are given

A bee flies at 20 feet per second directly to a flower bed from its hive

so, forward speed is

[tex] s_f=20 [/tex]

flies directly back to the hive at 12 feet per second

so, backward speed is

[tex] s_b=12 [/tex]

It is away from the hive fore 20 minutes total

so,

total time = forward time + backward time + stay time

The bee stays at the flower bed for 15 minutes,

20 =tf+tb+15

[tex] t_f+t_b=5min [/tex]

now, we can change it into seconds

[tex] t_f+t_b=5*60sec [/tex]

[tex] t_f+t_b=300sec [/tex]

(a)

we know that

distance between hive and flower bed will remain same

Let's assume

forward time =x

so,

[tex] t_f=x [/tex]

[tex] t_b=300-x [/tex]

we know that

distance = speed*time

[tex] d=s_f*t_f [/tex]

now, we can plug values

[tex] d=20*x [/tex]

[tex] d=s_b*t_b [/tex]

now, we can plug values

[tex] d=12*(300-x) [/tex]

both distances are same

[tex] 20x=12*(300-x) [/tex]

(b)

now, we can solve above equations and find x

[tex] 20x=12*(300-x) [/tex]

[tex] 20x=3600-12x [/tex]

add both sides 12x

[tex] 20x+12x=3600-12x+12x [/tex]

[tex] 32x=3600 [/tex]

[tex] x=112.5 [/tex]

now, we can find distance

[tex] d=20*112.5 [/tex]

[tex] d=2250 [/tex]

so, distance between hive and flower bed is 2250 feet.........Answer


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