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A particle is moveint along the x-axis whose instantaneous speed is given by `v^(2)=108-9x^(2)`. The acceleration of the particle is.A. `-9x ms^(-2)`B. `-18x ms^(-2)`C. `(-9x)/(2) ms^(-2)`D. None of there |
Answer» Correct Answer - A `v^(2)=18-9x^(2)` `a=(dv)/(dt)=(dv)/(dx).(dx)/(dt) =(d(sqrt108-9x^(2)))/(dx).(dx)/(dt)` `a=(1(-18x))/(2sqrt(108-9x^(2))).sqrt(108-9x^(2)) =-9x^(-2)` Alternative: Differentiating w.r.t. we get `2v(dv)/(dx)=-18x` `(vdx)/(dx)=-9x rArr a=-9x because (v(dv)/(dx)=a)`. |
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