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Homework 03 Solutions

# Homework 03 Solutions - homework 03 GADHIA TEJAS Due Feb 7...

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homework 03 – GADHIA, TEJAS – Due: Feb 7 2007, 4:00 am 2 Question 5 part 1 of 2 10 points A particle has constant acceleration vector vector a with components a x = 2 . 2 m / s 2 and a y = 3 . 1 m / s 2 . Its initial velocity vector vector v 0 (at time t 0 = 0) has components v 0 x = 3 m / s and v 0 y = 2 . 5 m / s. What is the particle’s speed at time t = 3 . 3 s? Correct answer: 12 . 846 m / s (tolerance ± 1 %). Explanation: Basic Concepts: For a motion with a constant acceleration vector, the particle’s velocity vector changes with time according to vector v ( t ) = t × vector a + vector v 0 . (1) And speed is the magnitude of the velocity vector, v = bardbl vector v bardbl = radicalBig ( v x ) 2 + ( v y ) 2 . (2) Solution: In vector notations, the velocity of the par- ticle at time t is given by eq. (1). To evaluate this equation, it is easiest to switch to compo- nent notations, thus: v x = t × a x + v 0 x = (3 . 3 s) × (2 . 2 m / s 2 ) + 3 m / s = 10 . 26 m / s , (3) and v y = t × a y + v 0 y = (3 . 3 s) × ( 3 . 1 m / s 2 ) + 2 . 5 m / s = 7 . 73 m / s .
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Homework 03 Solutions - homework 03 GADHIA TEJAS Due Feb 7...

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