A driver accelerates when the car is traveling at a speed of 30 miles per hour (i.e., 44 feet per second). the velocity (in feet per second) function is v(t)=44+2.2t . the car reaches the speed of 60 miles per hour (i.e., 88 feet per second) in 20 seconds. then during the 20 seconds the car has traveled

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.###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.