Figure shows (x,t) (y,t) diagram of a particle moving in 2-dimensions.

If the particle has a mass of 500 g , find the force (direction and magnitude) acting on the particle .

Explanation

Clearly from diagram (a) , the variation can be related as
x = t `implies (dx)/(dt) = 1` m/s
`a_(x) = 0`
From diagram (b) `" " y = t^(2)`
`implies " " (dy)/(dt) = 2t` or `a_(y) = (d^(2)y)/(dt^(2)) = 2 m//s^(2)`
Hence , `" " I_(y) = ma_(y) = 500 xx 10^(-3) xx 2 = 1N " " (because m = 500 g)`
`F_(x) = ma_(x) = 0`
Hence , net force , `" " F = sqrt(F_(x)^(2) + F_(y)^(2)) = F_(y) = 1N " " `(along y-axis)