Answer: 4(−a+1)(a−1)
Step-by-step explanation:
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](https://tex.z-dn.net/?f=d%3D%5Cint%5Climits%20%7Bv%7D%20%5C%2C%20dt%5C%5C%5C%5Cd%3D%5Cint%5Climits%20%7B%282i-4tj%29%7D%20%5C%2C%20dt%20%5C%5C%5C%5Cd%3D%282t%29i%2B%5Cdfrac%7B4t%5E2%7D%7B2%7Dj%5C%5C%5C%5Cd%3D%282t%29i%2B%282t%5E2%29j)
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](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bdv%7D%7Bdt%7D%5C%5C%5C%5Ca%3D%5Cdfrac%7B2i-4tj%7D%7Bdt%7D%5C%5C%5C%5Ca%3D%5Cdfrac%7B2i%7D%7Bdt%7D-%5Cdfrac%7B4tj%7D%7Bdt%7D%5C%5C%5C%5Ca%3D0-4j%5C%5C%5C%5Ca%3D-4j)
Hence , this is the required solution .
The difference is 180 is the correct answer for the problems