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There are several types of drag on a car other than air resistance. Effects having to do with the squeezing of the tires (rolling resistance) and frictional forces in the drivetrain (the system that transfers energy from the engine to the rotation of the wheels) also must be taken into account. Engineers use the following equation to model the total force due to these different effects Fdrag=A+Bv+Cv2 For a Accord, these coefficients are estimated to be A=220.500 N, B=−5.930 N s/m, and C=0.611 N s2/m2. Suppose that the driver steadily accelerates the car from 0 km/hr to 100 km/hr over a 3.5 s. What is the magnitude of the work done by the drag forces?]

Respuesta :

Answer:[tex]W=16.837\ kJ[/tex]

Explanation:

Given

Drag force is given by

[tex]F_{drag}=A+Bv+Cv^2[/tex]

for [tex]A=220.5\ N[/tex]

[tex]B=-5.93\ N-s/m[/tex]

[tex]C=0.611N-s/m^2[/tex]

car accelerate from 0 to [tex]100\ km/hr[/tex] in [tex]3.5\ s[/tex]

so acceleration is given by

[tex]v=u+at[/tex]

here u=initial velocity is zero

[tex]v=100\km/hr\approx 27.78\ m/s[/tex]

[tex]27.78=0+a(3.5)[/tex]

[tex]a=7.936\ m/s^2[/tex]

Now work done is given by

[tex]dW=F\cdot vdt[/tex]

[tex]\int_{0}^{W}dW=\int_{0}^{3.5}F\cdot vdt[/tex]

[tex]W=\int_{0}^{3.5}[Av+Bv^2+Cv^3]dt[/tex]

[tex]W=\int_{0}^{3.5}[220.5at-5.93a^2t^2+0.611a^3t^3]dt[/tex]

[tex]W=\int_{0}^{3.5}220.5\times 7.936tdt-\int_{0}^{3.5}5.93\times (7.936)^2t^2dt+\int_{0}^{3.5}0.611\times (7.936)^3t^3dt[/tex]

[tex]W=1749.88[\frac{t^2}{2}]_0^{3.5}-373.47[\frac{t^3}{3}]_0^{3.5}+305.389[\frac{t^4}{4}]_0^{3.5}[/tex]

[tex]W=10,718.015-5337.508+11,456.85[/tex]

[tex]W=16.837\ kJ[/tex]

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