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 12.9m/s². I'm certain that the maximum velocity is V(t → ∞) = 90.25m/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

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

$$s=\displaystyle\int_{0}^{9.675}v(t)dt=\displaystyle\int_{0}^{9.675}90.25(1-e^{-t/7})dt=$$ $$=90.25\bigg[t+7e^{-t/7}\bigg]\begin{matrix} 9.675\\ 0 \end{matrix}=$$ $$=90.25(9.675+7e^{-9.675/7}-0-7)\approx400.01 (m)$$

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

Check the work of the model amd compare the results

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

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

Time of reach 400m

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

$$\delta={|9.675-10.46| \over 10.46}\cdot100\%\approx7.5\%$$

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