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The x and y coordinates of a particle at any time t are given by `x=7t+4t^2` and `y=5t`, where x and t is seconds. The acceleration of particle at `t=5`s isA. zeroB. `8 ms^(-2)`C. `20 ms^(-2)`D. `40 ms^(-2)` |
Answer» Correct Answer - A `a=sqrt(a_(x)^(2)+a_(y)^(2))=[((d^(2)x)/(dt^(2)))^(2)+((d^(2)y)/(dt^(2)))^(2)]^(1//2)` Here `(d^(2)y)/(dt^(2))=0` Hence, `a=(d^(2)x)/(dt^(2))=8 ms^(-2)` |
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