Question

The position of an object in a video game is represented by an ordered pair. The coordinates of the ordered pair give the

Answer 1

Answer:

a. 457 pixels b. 229 pixels c. 114 pixels

Step-by-step explanation:

a. Using d = √[(x₂ - x₁)² + (y₂ - y₁)²] to find the distance apart of the coordinates (x₁, y₁) and (x₂, y₂) where (x₁, y₁) = (36, 315) and (x₂, y₂) = (410, 53).

So, d = √[(x₂ - x₁)² + (y₂ - y₁)²]

= √[(410 - 36)² + (53 - 315)²]

= √[374² + 262²]

= √[139,876 + 68,644]

= √208,520

= 456.64

= 457 pixels to the nearest pixel.

b. Since the players are d' distance apart, and moving at a same speed of v, if they approach each other, they meet at time t. So player A covers a distance of d = vt. For them to meet, player B approaches and covers a distance of d = d' - vt. So they meet when vt = d' - vt.

Solving this, we have

vt = d' - vt

vt + vt = d'

2vt = d'

t = d'/2v

Substituting t into d = vt, we have

d = vt

= v(d'/2v)

= d'/2

= 457/2

= 228.5

≅ 229 pixels  to the nearest pixel

c. Since the players are d' distance apart, and player A moves at a speed three times that of player B, if player B moves with speed v, then player A moves with speed 3v, as they approach each other, they meet at time t. So player A covers a distance of d = 3vt. For them to meet, player B approaches and covers a distance of d = d' - vt. So they meet when 3vt = d' - vt.

Solving this, we have

3vt = d' - vt

3vt + vt = d'

4vt = d'

t = d'/4v

Substituting t into d = vt, we have

d = vt

= v(d'/4v)

= d'/4

= 457/4

= 114.25

≅ 114 pixels to the nearest pixel