A particle moving with a velocity equal to `0.4 m//s` is subjected to an acceleration of `0.15 m//s^(2)` for `2 s`. in a direction at the right angle to its direction of motion. The resultant velocity is

A particle moving with a velocity equal to `0.4 m//s` is subjected to

1.	A particle moving with a velocity equal to `0.4 m//s` is subjected to an acceleration of `0.15 m//s^(2)` for `2 s`. in a direction at the right angle to its direction of motion. The resultant velocity is
Answer» In vector form, 1 st equation of motion is `vec(v) = vec(u) + vec(a)t` So, `v = sqrt(u^(2) + (at)^(2) + 2u (at) cos theta)` Here `u = 0.4 m//s, a = 0.15 m//s^(2), t = 2s` and `theta = 90^(@)` So, `u = sqrt([(0.4)^(2) + (0.15 xx 2)^(2) + 0]) = 0.5 m//s`

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A particle moving with a velocity equal to `0.4 m//s` is subjected to an acceleration of `0.15 m//s^(2)` for `2 s`. in a direction at the right angle to its direction of motion. The resultant velocity is

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