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Answers
Step-by-step explanation:
Let velocity of car = a
Let velocity of car = aThan a vector = ai = 20i (given)
Let velocity of car = aThan a vector = ai = 20i (given)Where i is unit vector along east word direction
Let velocity of car = aThan a vector = ai = 20i (given)Where i is unit vector along east word directionAnd let velocity of bus is = b vect
Let velocity of car = aThan a vector = ai = 20i (given)Where i is unit vector along east word directionAnd let velocity of bus is = b vectSo b vec - a vec = 20root3 j
Let velocity of car = aThan a vector = ai = 20i (given)Where i is unit vector along east word directionAnd let velocity of bus is = b vectSo b vec - a vec = 20root3 j(given, here j because it's along North which is y asis)
Let velocity of car = aThan a vector = ai = 20i (given)Where i is unit vector along east word directionAnd let velocity of bus is = b vectSo b vec - a vec = 20root3 j(given, here j because it's along North which is y asis)So b = 20root 3 j +20i
Let velocity of car = aThan a vector = ai = 20i (given)Where i is unit vector along east word directionAnd let velocity of bus is = b vectSo b vec - a vec = 20root3 j(given, here j because it's along North which is y asis)So b = 20root 3 j +20iMode of b = root ( 20*20*3 + 20*20) =root 1600 = 40 m/s
Let velocity of car = aThan a vector = ai = 20i (given)Where i is unit vector along east word directionAnd let velocity of bus is = b vectSo b vec - a vec = 20root3 j(given, here j because it's along North which is y asis)So b = 20root 3 j +20iMode of b = root ( 20*20*3 + 20*20) =root 1600 = 40 m/sAnd tan theta = 20root3/20=root 3
Let velocity of car = aThan a vector = ai = 20i (given)Where i is unit vector along east word directionAnd let velocity of bus is = b vectSo b vec - a vec = 20root3 j(given, here j because it's along North which is y asis)So b = 20root 3 j +20iMode of b = root ( 20*20*3 + 20*20) =root 1600 = 40 m/sAnd tan theta = 20root3/20=root 3Hence angle = 60 degree towards East to north
Answer:
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