Question

Figure shows (vx, t) and (vy, t) diagrams for a body of unit mass. Find the force as a function of time.

(a)

(b)

Answer 1

In figure (a), we have v_x = 2t for t < 1s and v_x = 2(2 – t) for 1s < t < 2s

Hence the acceleration between 0s to 1s will be given by,

⇒ `a_x = (dv_x)/(dt) = (d^2t)/(dt) = 2  ms^-2`

Also `F_x = ma_x = 1 xx 2 = 2N` for t < 1s

And the acceleration between 1s to 2s will be given by,

⇒ `a_x = (dv_x)/(dt) = (2d(2 - t))/(dt) = - 2  ms^-2`

Also `F_x = ma_x = 1 xx (-2) = - 2N` for 1s < t < 2s

Now in figure (b), we have v_y = t for t < 1s and v_y = 1 for t > 1s

Hence the acceleration between 0s to 1s will be given by,

⇒ `a_y = (dv_y)/(dt) = (dt)/(dt) = 1 ms^-2`

Also `F_y = ma_y = 1 xx 1 = 1N` for t < 1s

And the acceleration between 1s to 3s will be given by,

⇒ `a_y = (dv_y)/(dt) = (d1)/(dt) = 0 ms^-2`

Also `F_y = ma_y = 1 xx 0 = 0N` for t > 1s

Now the resultant force will be given by,

⇒ `vecF = vec(F)_xhati + vec(F)_yhatj`

Hence,

⇒ `vecF = 2hati + hatj` for t < 1s

And,

⇒ `vecF = - 2hati` for 1s < t < 2s

And,

⇒ `vecF = 0` for t > 2s