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
