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

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 ... 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

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

karton

$\frac{dy\left(t\right)}{dt}=10t$
y(t) is distance travelled in time t
Now t vares from 0 to 4 sec
then
$=\left[\frac{10{t}^{2}}{2}{\right]}_{0}^{4}$
$=\left[5{t}^{2}{\right]}_{0}^{4}=80-0$

Vasquez

80 meters
Explanation:
We are given that the car's speed is $10t$ m/s $t$ seconds after starting. This can be written as:
$v\left(t\right)=10t$
The distance traveled by the car can be obtained by integrating the car's speed with respect to time. Since speed is the derivative of distance with respect to time, we have:
$\frac{dx}{dt}=v\left(t\right)$
Integrating both sides of the equation with respect to $t$, we get:
$\int dx=\int v\left(t\right)dt$
$x=\int 10tdt$
Integrating, we find:
$x=5{t}^{2}+C$
where $C$ is the constant of integration. Since the car starts from a stop, we know that at $t=0$, the distance traveled is zero. Therefore, we can substitute $t=0$ and $x=0$ into the equation to find the value of $C$:
$0=5\left(0{\right)}^{2}+C$
$C=0$
Substituting this value back into the equation, we have:
$x=5{t}^{2}$
Now we can calculate the distance traveled by the car in 4 seconds by substituting $t=4$ into the equation:
$x=5\left(4{\right)}^{2}$
$x=80$
Therefore, the car will travel 80 meters in 4 seconds.

RizerMix

To solve the problem using differential equations, let's first establish the relationship between the car's speed and time. The problem states that the car's speed is given by $10t$ m/s at time $t$ seconds after starting. We can express this as a first-order linear ordinary differential equation:
$\frac{ds}{dt}=10t$
To find the distance the car will travel in 4 seconds, we need to integrate the equation. Integrating both sides with respect to $t$ gives:
$\int ds=\int 10t\phantom{\rule{0.167em}{0ex}}dt$
The left-hand side becomes the distance traveled $s$, and the right-hand side is the integral of $10t$, which is $\frac{1}{2}{t}^{2}+C$, where $C$ is the constant of integration. Thus, we have:
$s=\frac{1}{2}{t}^{2}+C$
To determine the value of the constant $C$, we can use the initial condition that the car starts from a stop. At $t=0$, the speed is 0 m/s, which implies that the initial distance traveled is also 0. Substituting these values into the equation, we have:
$0=\frac{1}{2}\left(0{\right)}^{2}+C$
$0=0+C$
$C=0$
Therefore, the equation describing the distance traveled by the car in terms of time is:
$s=\frac{1}{2}{t}^{2}$
Now, let's find the distance the car will go in 4 seconds. Substituting $t=4$ into the equation, we have:
$s=\frac{1}{2}\left(4{\right)}^{2}$
$s=8$
Hence, the car will travel 8 meters in 4 seconds.

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