InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
If the time and displacement of the particle along the positive x-axis are related as `t=(x^(2)-1)^(1//2)`, find the acceleration in terms of `x`. |
|
Answer» `t=(x^(2)-1)^(1//2)` `(dt)/(dx)=1/2(x^(2)-1)^(1//2)(2x)=x/((x^(2)-1)^(1//2))` `v=((x^(2)-1)^(1//2))/x=(1-1/x^(2))^(1//2)` `(dv)/(dx)=1/2(1-1/x^(2))^(-1//2)(2/x^(3))` `a=v(dv)/(dx)=1/x^(3)` OR `t=(x^(2)-1)^(1//2)impliesx=(1+t^(2))^(1//2)` `v=(dx)/(dt)=1/2(1+t^(2))^(-1//2)(2t)` `=t/((1+t^(2))^(1//2))=1/((1/t^(2)+1)^(1//2))=(1+1/t^(2))^(-1//2)` `a=(dv)/(dt)=-1/2(1+1/t^(2))^(-3//2)(-2/t^(3))` `=1/(t^(3)(1+1/t^(2))^(3//2))=1/((t^(2)+1)^(3))` `=1/x^(3)` OR `t=(x^(2) -1)^(1//2)` `t^(2)=(x^(2)-1)` Differentiating w.r.t. t, we get `2t=2x(dx)/(dt)=2xv` `t=xv` Differentiating w.r.t. t we get `1=x(dv)/(dt)+v(dx)/(dt)=xa+v^(2)` (Product rule) `ax=1-v^(2)=1-(t^(2))/(x^(2))=(x^(2)-t^(2))/(x^(2))=(1)/(x^(2))` `a=(1)/(x^(3))` Though the above method is simple, but for this, you will require further knowledge of differentiation. e.g. if x and y are functions of `t` (not shown in expression), `y=x^(3)+3x^(2)` differentiating w.r.t. t, we get `(dy)/(dt)=3x^(2)(dx)/(dt)+3.2x(dx)/(dt)` Differentiating w.r.t. `t`, we get `2y(dy)/(dt)=3x^(2)(dx)/(dt)` We can also write `2y dy = 3x^(2) dx` |
|