Question

Andrew starts walking at noon at a speed of 3 miles per hour. Christian leaves the same place an hour later and walks in the same direction at a speed of 4 miles per hour. What time will Christian catch up with Andrew?

Answer 1

Answer:

Christian will catch up with Andrew at 4 PM.

Step-by-step explanation:

We use the equations of their positions to solve this question. This equation has the following format:

$S(t) = S(0) + at$

In which S(t) is the position after t hours, S(0) is the initial position, and a is their speed.

Andrew starts walking at noon, Christian at 1.

We can model the equations starting from 1.

Andrew:

Starts at noon, 3 mph.

So at 1, when Christian starts walking, he is at the mile 3. So S(0) = 3. 3mph, a = 3. So

$S_{a}(t) = 3 + 3t$

Christian:

Starts at 1, 4 mph.

So at 1, he is at the position 0. So S(0) = 0. 4 mph, so a = 4.

$S_{c}(t) = 4t$

What time will Christian catch up with Andrew?

t hours after Christian starts walking(which is 1 pm).

t is found when

$S_{c}(t) = S_{a}(t)$

$4t = 3 + 3t$

$t = 3$

3 hours after 1 PM, which is 4PM.

Christian will catch up with Andrew at 4 PM.