Assume the car starts at the origin, so that its initial position is [tex]x(0)=0[/tex]. The car's displacement at any time [tex]t[/tex] over the 20 second interval is
[tex]\displaystyle x(0)+\int_0^t(44+2.2u)\,\mathrm du=0+\left(44u+1.1u^2\right)\bigg|_{u=0}^{u=t}=44t+1.1t^2[/tex]
so that after 20 seconds the car has moved 1320 ft.
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Without using calculus, recall that under constant acceleration, the average velocity of the car over the 20 second interval satisfies
[tex]v_{\rm avg}=\dfrac{v_f+v_i}2[/tex]
and that, by definition, we have
[tex]v_{\rm avg}=\dfrac{\Delta x}{\Delta t}[/tex]
where [tex]v_f[/tex] and [tex]v_i[/tex] are the final/initial speeds of the car and [tex]\Delta x[/tex] is the displacement it undergoes. It starts with a speed of 44 ft/s and ends with a speed of 88 ft/s, so we have
[tex]\dfrac{88\frac{\rm ft}{\rm s}+44\frac{\rm ft}{\rm s}}2=\dfrac{\Delta x}{20\,\rm s}\implies\Delta x=1320\,\mathrm{ft}[/tex]
same as before.
