Answer:
Acceleration,
\[\text{a}=\frac{dv}{dt}=\frac{dv}{dx}\times \frac{dx}{dt}=\upsilon
\frac{dv}{dx}\]
Given,\[{{\upsilon }^{\text{2}}}=\left(
\text{18}0-\text{16x} \right)\]; Differentiating it w.r.t. x, we have \[\text{2}\upsilon
\frac{dv}{dx}=-\text{16}\]
So, acceleration, \[a=\upsilon
\frac{dv}{dx}=-\frac{16}{2}=-\text{8m}/{{\text{s}}^{\text{2}}}\]
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