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An object's position in a given coordinate system is described by the vector r = t2 i - (3t + 3) j. Assume the object moves without air resistance. einaaa@berkeley.edu 25%

Part (a) What is the object's displacement between times t1 = 1 s and t2 = 3 s? Write your answer in component form. Δr = 8 i - 6 j ✔ Correct!

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mauricioalessandrell

Answer:

The object´s displacement vector is Δr =  8i - 6 j  

Explanation:

Hi there!

The position vector is given by the following function:

r = t²i - (3t + 3) j

Let´s find the position of the object at time t1 and t2:

At t1 = 1 s:

r1 = (1)² i - (3 · (1) + 3 )j

r1 = 1 i - 6 j

At t2 = 3 s:

r2 = (3)² i - (3 · (3) + 3) j

r2 = 9 i - 12 j

The displacement is calculated as follows:

displacement = Δr = final position - initial position = r2 - r1

r2 - r1 = 9 i - 12 j - (1 i - 6 j)

r2 - r1 = 9 i - 12 j - 1 i + 6 j

r2 - r1 = 8 i - 6 j

The object´s displacement vector is Δr =  8i - 6 j  

onyebuchinnaji

The object's displacement for the given time period is 8i - 6j

The given parameter:

[tex]r = t^2 i - (3t+3)j[/tex]

t₁ = 1 s

t₂ = 3 s

To find:

  • the displacement of the object between t₁ and t₂

When the time is 1 s, the position of the object is calculated as;

[tex]r_1 = (1)^2i - (3\times 1 + 3)j\\\\r_1 = i - (6)j\\\\r_1 = i - 6j[/tex]

When the time is 3 s, the position of the object is calculated as;

[tex]r_2= (3)^2i - (3\times 3 + 3)j\\\\r_2 = 9i - (9+3)j\\\\r_2 = 9i - 12j[/tex]

The displacement of the object is calculated as;

[tex]\Delta r = r_2 -r_1\\\\\Delta r = 4i - 9j - (i - 6j)\\\\\Delta r = 9i-12j -i + 6j\\\\\Delta r = (9i -i) + (-12j + 6j)\\\\\Delta r = 8i - 6j[/tex]

Thus, the object's displacement for the given time period is 8i - 6j

Learn more here: https://brainly.com/question/9972743

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