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ollegr [7]
3 years ago
12

you and a friend of your choice are driving to the island of chincoteague in virginia in two different cars. you are traveling 6

5 miles per hour and your friend is traveling 51 miles per hour. your friend has a 35-mile head start. the island of cincoteague is about 200 miles from philadelhia. when will you catch up with your friend
Mathematics
1 answer:
andrey2020 [161]3 years ago
6 0

Answer:

<em>They will catch up at 2.5 hours</em>

Step-by-step explanation:

Constant Speed Motion

It refers to situations where objects move at the same speed in the same direction. The speed is computed as the ratio of the distance x to the time

\displaystyle v=\frac{x}{t}

The situation pictured in the question requires the computation of distances traveled by two objects running at different speeds. The slower car goes ahead with a 35-mile head start, so it's just a matter of time when the fastest car and the slower will catch up. We'll use here the concept of relative speed to easily solve the problem.

The speed of the first friend is

v_1=51\ mph

And the speed of the second friend is

v_2=65\ mph

If the first friend was at rest, the second friend will have a relative speed of

v_r=65-51=14\ mph

The distance between them (35 miles) will be covered in a time t which can be found from

\displaystyle v_r=\frac{x}{t}

\displaystyle t=\frac{x}{v_r}

\displaystyle t=\frac{35}{14}=2.5\ hours

We need to check if they have not arrived to chincoteague by that time. We'll compute the distance of the fastest friend

x_2=65*2.5=162.5\ miles

We can see they catch up (200-162.5=37.6 miles) before arriving to their destiny.

Just to verify, the distance traveled by the first friend is

x_2=51*2.5=127.5\ miles

The difference of those distances is 162.5-127.5=35 miles, exactly the original head-start

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