09J-Further-Apps


 * Further Applications of Integration **


 * Example 1 **

The rate of change of position (velocity) of a particle travelling in a straight line is given by math . \qquad \dfrac{dx}{dt}=40-10e^{-0.4t} \;\; m/s, \;\; t \geqslant 0 \qquad. math where x is measured in metres and t in seconds.


 * a) Find the initial velocity **


 * This occurs when t = 0. **

math \\ . \qquad v(0) = 40-10e^0 \qquad. \\ .\\ . \qquad \qquad = 40 - 10 \qquad. \\ . \\ . \qquad \qquad = 30 \textit{ m/s} \qquad. math


 * b) Find the velocity after 10 seconds (correct to 2 decimal places) **

math \\ . \qquad v(10) = 40-10e^{-0.4 \times 10} \qquad. \\ .\\ . \qquad \qquad = 40 - 10e^{-4} \qquad. \\ . \\ . \qquad \qquad = 39.82 \textit{ m/s} \qquad. math


 * c) Find the time taken to reach a velocity of 35 m/s **

math \\ . \qquad v(t) = 35 \qquad .\\. \\ . \qquad 40-10e^{-0.4t} = 35 \qquad. \\ . \\ . \qquad -10e^{-0.4t} = -5 \qquad. \\ .\\ . \qquad e^{-0.4t} = 0.5 \qquad. \\ . \\ . \qquad -0.4t = \log_e(0.5) \qquad. \\ .\\ . \qquad t = \dfrac{\log_e(0.5)}{-0.4} \qquad. \\ . \\ . \qquad t = 1.73 \textit{ seconds} \qquad. math


 * d) Sketch the graph of dx/dt against t **


 * e) Find the total distance travelled by the particle in the first 10 seconds. **

math . \qquad \dfrac{dx}{dt}=40-10e^{-0.4t} \qquad. math


 * Particle doesn't change directions, so distance travelled equals the area under the graph **

math . \qquad s = \displaystyle{\int}_0^{10} 40 - 10e^{-0.4t} \; dt \qquad. math

math . \qquad s = \Big[ 40t + 25e^{-0.4t} \Big]_0^{10} \qquad. math

math \\ . \qquad \; = \big( 400 + 25e^{-4} \big) - \big( 25 \big) \qquad. \\ .\\ . \qquad \; = 375 + 25e^{-4} \qquad. \\ . \\ . \qquad \; = 375.46 \textit{ m} \qquad. math .