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Nezavi [6.7K]
3 years ago
11

Show the working for 32.2x62.1

Mathematics
1 answer:
Margaret [11]3 years ago
8 0
32.2 x 62.1=1999.62 i cant show my work but i know the ans
 
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A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v>=2i-4tj^
guapka [62]

Given :

A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v=2i-4tj  .

To Find :

A. The vector position of the particle at any time t .

B. The acceleration of the particle at any time t .

Solution :

A )

Position of vector v is given by :

d=\int\limits {v} \, dt\\\\d=\int\limits {(2i-4tj)} \, dt \\\\d=(2t)i+\dfrac{4t^2}{2}j\\\\d=(2t)i+(2t^2)j

B )

Acceleration a is given by :

a=\dfrac{dv}{dt}\\\\a=\dfrac{2i-4tj}{dt}\\\\a=\dfrac{2i}{dt}-\dfrac{4tj}{dt}\\\\a=0-4j\\\\a=-4j

Hence , this is the required solution .

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The answer to the question

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Answer:

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Step-by-step explanation:

28 \leq 4n

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