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The position vector of a particle at time \(t\) is given by \(\mathbf{r}(t)=n e^{-2 t}\,\mathbf{i}-t^2\,\mathbf{j}\), where \(n\) is a positive constant.
Determine \(n\) if the particle's acceleration is perpendicular to its velocity when \(t=\dfrac{1}{2}\). (3 marks)
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\(n=\dfrac{e}{2}\)
| \(\underset{\sim}{r}(t)\) | \(=n e^{-2 t} \underset{\sim}{i}-t^2 \underset{\sim}{j}\) |
| \(\underset{\sim}{v}(t)\) | \(=-2n e^{-2t} \underset{\sim}{i}-2 t\underset{\sim}{i} \ \ \Rightarrow \ \ v\left(\frac{1}{2}\right)=-2 ne^{-1} \underset{\sim}{i}-\underset{\sim}{j}\) |
| \(\underset{\sim}{a}(t)\) | \(=4 ne^{-2t} \underset{\sim}{i}-2 \underset{\sim}{j} \ \ \Rightarrow \ \ a\left(\frac{1}{2}\right)=4ne^{-1} \underset{\sim}{i}-2 \underset{\sim}{j}\) |
\(\text{Velocity}\perp \text{acceleration at}\ \ t=\dfrac{1}{2}:\)
\(\displaystyle \binom{-\tfrac{2 n}{e}}{-1}\binom{\tfrac{4 n}{e}}{-2}=0\)
\(-\dfrac{8 n^2}{e^2}+2=0 \ \ \Rightarrow \ \ n^2=\dfrac{e^2}{4} \ \ \Rightarrow \ \ n=\dfrac{e}{2}\ \ (n\gt 0)\)