A particle moves in x-y plane according to equations x=4t^2+5t+16 and 6y=5t. What is the acceleration of the particle?

A particle moves in x-y plane according to equations x=4t^2+5t+16 and

1.	A particle moves in x-y plane according to equations x=4t^2+5t+16 and 6y=5t. What is the acceleration of the particle?
Answer» Solution :`x = 4t^2+5T+16` y= 5t `v_x = dx/DT = 8t+5` v_y =5 `a_x = (dv_x)/dt = 8` (dv_y)/dt = a_y = 0 `a = SQRT(a_x^2+a_y^2) = sqrt(8^2+0)` = 8m / `s^2`

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A particle moves in x-y plane according to equations x=4t^2+5t+16 and 6y=5t. What is the acceleration of the particle?

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