08 7apps

# Further Applications of Integration

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### Example 1

The velocity (rate of change of position) of a particle travelling in a straight line is given by

… … $v(t) = \dfrac{dx}{dt}=40-10e^{-0.4t} \;\; m/s, \;\; t \geqslant 0$

… … … where x is measured in metres and t in seconds.

a) .. Find the initial velocity

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

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

d) .. Sketch the graph of $\dfrac{dx}{dt}$ against t

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

a) .. Find the initial velocity

… … … Initial velocity occurs when $t = 0$.

… … … $v(t) = \dfrac{dx}{dt}=40-10e^{-0.4t}$

… … … $v(0) = 40 - 10e^0$

… … … … … $= 40 - 10$

… … … … … $= 30$ m/s
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b) .. Find the velocity after 10 seconds (correct to 2 decimal places)

… … … $v(t) = 40-10e^{-0.4t}$

… … … $v(10) = 40-10e^{-0.4 \times 10}$

… … … … … $= 40 - 10e^{-4}$

… … … … … $= 39.82$ m/s
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c) .. Find the time taken to reach a velocity of 35 m/s

… … … $v(t) = =40-10e^{-0.4t}$

… … … Let $v(t) = 35$

… … … $40-10e^{-0.4t} = 35$

… … … $-10e^{-0.4t} = -5$

… … … … $e^{-0.4t} = 0.5$

… … … $-0.4t = \log_e(0.5)$

… … … $t = \dfrac{\log_e(0.5)}{-0.4}$

… … … $t = 1.73$ seconds

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d) .. Sketch the graph of $\dfrac{dx}{dt}$ against t

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

… … … $\dfrac{dx}{dt}=40-10e^{-0.4t}$
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… … … Particle doesn't change directions,
… … … so distance travelled equals the area under the graph

… … … $s = \displaystyle{\int}_0^{10} 40 - 10e^{-0.4t} \; dt$
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… … … $s = \Big[ 40t + 25e^{-0.4t} \Big]_0^{10}$
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… … … .. $= \Big( 400 + 25e^{-4} \Big) - \Big( 25 \Big)$
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… … … .. $= 375 + 25e^{-4}$
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… … … .. $= 375.46$ m

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