Answer to Question #105088 in Calculus for Leon

Question

Given data is that car has a model that states use equation (7 at top 0 and bottom of integral) 7^0∫v(t)dt, 0-28m/s of 2.6s, time of reach 400m of 10.46s and tmaxspeed of 7s. I have calculated that value of coefficient A is 90.25m/s and time to reach max acceleration is 23.7m/s². I'm certain that the max velocity is at t=7 v(7)=90.25 x (1-e^-10 = 57.05m/s). Please draw a graph for velocity v time showing velocity m/s on y axis also labelled with A, max velocity/speed. Then on the x axis please show where tmax is hit and where t400 is hit with x axis labelled time (s) please also show Vmax along the graph. once this is done please work out the areas for both sections and write a brief explanatation on the comparison between the model and my results :)

Answer 1

$v(t)=90.25(1-e^{-t/7})$ $v(2.6)=90.25(1-e^{-2.6/7})\approx28 (m/s)$ $v(7)=90.25(1-e^{-7/7})\approx57 (m/s)$

$s=\displaystyle\int_{0}^{9.675}v(t)dt=\displaystyle\int_{0}^{9.675}90.25(1-e^{-t/7})dt=$ $=90.25\bigg[t+7e^{-t/7}\bigg]\begin{matrix} 9.675\\ 0 \end{matrix}=90.25(9.675+7e^{-9.675/7}-0-7)\approx$ $=400.01 (m)$

$v(9.675)=90.25(1-e^{-9.675/7})\approx67.59 (m/s)$

$rectangle: 67.59\ m/s \times\ 9.675 \ s=653.93\ m$

Check the work of the model and compare results

$v(2.6)=90.25(1-e^{-2.6/7})\approx28 (m/s),\ True$

$v(7)=90.25(1-e^{-7/7})\approx57 (m/s)$

Time of reach 400m

$t_{400}=9.675 \ s<10.46 \ s$

$\delta={|9.675-10.46| \over 10.46}\cdot100\%\approx 7.5\%$

I think based on the results, a model can be adopted to solve the problem.