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A particle has initial velocity (0.3i^+0.4j^)and has acceleration (0.1i^+0.2j^),it's magnitude of displacement after 10seconds is

A particle has initial velocity (0.3i^+0.4j^)and has acceleration (0.1i^+0.2j^),it's magnitude of displacement after 10seconds is
 

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Grade:12th pass

2 Answers

Saurabh Koranglekar
askIITians Faculty 10341 Points
one year ago
Dear student

Lets calculate for x part

S i^ = ux t + 0.5 ax t*t = 3 + 5= 8i

S j^ = uy t + 0.5 ay t*t = 4 + 10 = 14j

S = sqrt ( 260)

Regards
Khimraj
3007 Points
one year ago
V = u + at = 3i +4j + 10(0.3i +0.4j) 
= 3i +4j +4i + 3j
= 7i +7j

so magnitude of final velocity is √(7²+7²)= 7√2

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