abreviatsjw

2021-12-27

A racer accelerates from a stop so that its speed is 10t m/s t second after starting how far will the car go in 4 seconds using differential equations?

einfachmoipf

Beginner2021-12-28Added 32 answers

Explanation: Given $\frac{dy\left(t\right)}{dt}=10t$ where y(t) is the distance travelled a function of time above equation is a first order first degree DE where t varies from 0 to 4 seconds integrating on both side w.r.t we get $\int}_{0}^{4}\text{}dy\left(t\right)={\int}_{0}^{4}\text{}10t\text{}dt=y\left(4\right)-y\left(0\right)={\left[5{t}^{2}\right]}_{0}^{4$ ... but $y\left(0\right)=0$ since car is at rest at time $t=0$

$y\left(4\right)=5\left(16\right)=80m$ .

xandir307dc

Beginner2021-12-29Added 35 answers

Initial velocity is 0 and and final is 10 time is 4 sec then $\text{distance}=\frac{\text{average velocity}}{\text{time}}$

karton

Expert2022-01-09Added 613 answers

y(t) is distance travelled in time t

Now t vares from 0 to 4 sec

then

Vasquez

Expert2023-05-14Added 669 answers

RizerMix

Expert2023-05-14Added 656 answers

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I Completed the square on the bottom but what do you do now?

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$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$