A boy on a treasure hunt walks 12 paces West, then 21 paces North, then 25 paces East. What is the magnitude of boy’s resultant displacement in meters?

The displacement is 18.8 meters.

Explanation:

Let's define the North as the positive y-axis and the East as the positive x-axis.

We also can assume that if the point (x, y) represents the position of the boy, the initial position is (0 paces, 0 paces).

"A boy on a treasure hunt walks 12 paces West,"

West would be the negative x-axis in this case, then:

the position will be (-12 paces, 0 paces)

"then 21 paces North"

Now the position will be (-12 paces, 21 paces)

"then 25 paces East."

Now the position will be (-12 paces + 25 paces, 21 paces)

                                       = (13 paces, 21 paces)

Now we have the initial and final positions of the boy,  (0 paces, 0 paces) and (13 paces, 21 paces) respectively.

We want to find the displacement, which is defined as the distance between the final position and the initial position.

Remember that for two vectors (a, b) and (c, d), the distance is:

D = √( (a - c)^2 + (b  - d)^2)

In this case, the displacement will be:

D = √( ( 13 paces - 0 paces)^2 + (21 paces - 0 paces)^2) = 24.7 paces.

But we want this quantity in meters, so we can use the relation:

1 pace = 0.762 meters.

Then 24.7 paces, are 24.7 times 0.762 meters.

D = 24.7*0.762 m = 18.8 meters.

with 5 seconds, no, but over time if the police car stays at 65m/s ad the speeding car stays at 60, then obviously yes, but the answer to your question would be, no.

hope this : )