Question

A turtle and a snail agreed to have a race. Being kind, the turtle gave the snail a head start. The snail took off at a blinding 3 feet per minute. 50 minutes later, the turtle went after him at a blistering 9 feet per minute

S(m) = 150 + 3m models the snail's distance after the turtle started, where m is the minutes.

T (m) = 9m models the turtle's distance.

In how many minutes will the turtle catch up to the snail?

A. 25 minutes
B. 35 minutes
C. 30 minutes
D. 45 minutes

Answer 1

In 25 minutes turtle will catch up to the snail.

Explanation:

we have given,

Snail's speed = 3 feet per minute

Snail's had a head start , so after 50 minutes Snail's distance = 150

Now after 50minute we can model snail's distance equation for m minutes as:

S(m) = 150 + 3m

next, turtle went after him after 50 minutes at a speed of 9 feet per minute, and the distance turtle can cover in m minutes will be:

Turtle's distance, T(m) = 9m

We need to calculate the minutes when turtle catch up to the snail.

i.e distance covered by snails and turtle will equal at the point where they meet ,

S(m) = T(m)

or 150 + 3m = 9m

or 150 = 9m - 3m

or 150 = 6m

or m =
`(150)/(6) =25` minutes

So, In 25 minutes turtle will catch up to the snail.