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A particle is moving east wards with a velocity of 5 ms^(-1)In 10 S , the velocity changes to 5 ms^(-1)north wards. The average acceleration in this time is |
Answer» Solution :`vec(a)_(a v) = (vec(DELTA v))/(Delta t) = (vec(v)_(f) - vec(v)_(i))/(Delta t)` `vec(a)_(a v) = (5hat(j) - 5hat(j))/(10) = (5sqrt(2))/(10) = (1)/(sqrt(2))m//s^(2)` north-west direction. |
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