Question

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

Answer 1

Assume the car starts at the origin, so that its initial position is $x(0)=0$. The car's displacement at any time $t$ over the 20 second interval is

$\displaystyle x(0)+\int_0^t(44+2.2u)\,\mathrm du=0+\left(44u+1.1u^2\right)\bigg|_{u=0}^{u=t}=44t+1.1t^2$

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

$v_{\rm avg}=\dfrac{v_f+v_i}2$

and that, by definition, we have

$v_{\rm avg}=\dfrac{\Delta x}{\Delta t}$

where $v_f$ and $v_i$ are the final/initial speeds of the car and $\Delta x$ 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

$\dfrac{88\frac{\rm ft}{\rm s}+44\frac{\rm ft}{\rm s}}2=\dfrac{\Delta x}{20\,\rm s}\implies\Delta x=1320\,\mathrm{ft}$

same as before.