Let your character move in a constant acceleration (a + b / 2, -a-b) along the (a,b) direction with a velocity of (a+b, a-b). If the initial position of the character is at (a2b2, ab), where will the character be located after 2a + 3b + 4ab time units.  a = 5 b = 2 Solve using the following equation below

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Chapter1: Units, Trigonometry. And Vectors
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Let your character move in a constant acceleration (a + b / 2, -a-b) along the (a,b) direction with a velocity of (a+b, a-b). If the initial position of the character is at (a2b2, ab), where will the character be located after 2a + 3b + 4ab time units. 

a = 5

b = 2

Solve using the following equation below

P₁ = P₁-1 + V₁-1At +ä(At)²
2
Transcribed Image Text:P₁ = P₁-1 + V₁-1At +ä(At)² 2
Expert Solution
Step 1

given data,

a =5

b = 2

constant acceleration = (a + b/2), -a-b)

velocity = (a+b, a-b)

initial position = (a^2b^2, ab) 

we have to find the position after 56 units 

using given equation

so,

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