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The position vector of a particle changes with time according to the relation r(t) = 15t² î + (4 − 20t²) ĵ. What is the magnitude of the acceleration at t = 1 ?

1. 100
2. 40
3. 50
4. 25

Correct answer: 50

Solution

r=15t^2 i +(4-20t^2) j gives a = r'' = 30 i - 40 j (constant). |a| = sqrt(30^2+40^2) = sqrt(2500) = 50.

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