Integrating a velocity made of reciprocal-linear and exponential terms to find the displacement from a start at the origin
Integrating a velocity made of reciprocal-linear and exponential terms to find the displacement from a start at the origin
5 marks
A small robot moves in a straight line along a corridor. Its displacement from a fixed point \(O\) in the corridor is \(s\) metres, and its velocity \(t\) seconds after it starts is \(v\text{ m s}^{-1}\), where
\(v=\dfrac{4}{2t+1}-2\mathrm{e}^{-2t},\quad t\geq0.\)
- Find \(\displaystyle\int\left(\dfrac{4}{2t+1}-2\mathrm{e}^{-2t}\right)\mathrm{d}t\).[2]
The robot starts at \(O\).
- Express \(s\) in terms of \(t\).[3]
Solution:
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Answer:(a) \(2\ln(2t+1)+\mathrm{e}^{-2t}+C\) (b) \(s=2\ln(2t+1)+\mathrm{e}^{-2t}-1\)
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