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bulgar [2K]
3 years ago
10

Two dads and two son's went to an ice cream shop and they all bought one each it cost's 50 cents but they only paid $1.50

Mathematics
2 answers:
Burka [1]3 years ago
7 0

There were only 3 men.

One was the grandfather aka the father #1

One was the father to the child and simultaneously the son of the child's grandfather (father #2 and son #1)

One was the child (child #2)


Gnom [1K]3 years ago
6 0

If there were two dad's and two sons, and they each bought an ice cream cone worth 50, but only paid 1.50, that means there was a grandpa, who is a dad,

a dad, who is a dad of the child and a son of the grandpa,

and a son, who is a son of the dad.

This all adds up to two dads and two sons.

(the dad is both a dad and a son, if you don't understand ;))

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Answer:

a. 18.34 s b. 327.92 m

Step-by-step explanation:

a. How long before the police car catches the speeder who continued traveling at 40 miles/hour

The acceleration of the car a in 10 s from 0 to 55 mi/h is a = (v - u)/t where u = initial velocity = 0 m/s, v = final velocity = 55 mi/h = 55 × 1609 m/3600 s = 24.58 m/s and t = time = 10 s.

So, a =  (v - u)/t =  (24.58 m/s - 0 m/s)/10 s = 24.58 m/s ÷ 10 s = 2.458 m/s².

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s = 0 m/s × 10 s + 1/2 × 2.458 m/s² (10)²

s = 0 m + 1/2 × 2.458 m/s² × 100 s²

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The total distance moved by the police car is thus S = s + s' = 122.9 + 24.58t

The total distance moved by the speeder is S' = 40t' mi = (40 × 1609 m/3600 s)t' =  17.88t' m where t' = time taken for police to catch up with speeder.

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S' = S

17.88t' = 122.9 + 24.58t

Also, the time  taken for the police car to catch up with the speeder, t' = time taken for car to accelerate to 55 mi/h + rest of time taken for police car to catch up with speed, t

t' = 10 + t

So, substituting t' into the equation, we have

17.88t' = 122.9 + 24.58t

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So, it will take 18.34 s before the police car catches the speeder who continued traveling at 40 miles/hour

b. how far before the police car catches the speeder who continued traveling at 40 miles/hour

Since the distance moved by the police car also equals the distance moved by the speeder, how far the police car will move before he catches the speeder is given by S' = 17.88t' = 17.88 × 18.34 s = 327.92 m

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