A car, moving at a certain velocity and with constant acceleration, is applied brakes to make it stop. The car stops 10 seconds after the brakes are applied an travels 300 m during this time. Find the equation (ODE) of motion of the car during this 10 seconds interval. Moreover, determine distance covered, the speed, acceleration of the car, at the moment brakes are applied.
I have been trying to generate an ODE corresponding to the above situation, but I am unable to do so. I am not able to determine the ODE for this scenario and solve it.
Here is my (wrong) attempt to model this situation into an ODE:
$$\frac{\mathrm d^2x}{\mathrm dt^2} + 10\frac{\mathrm dx}{\mathrm dt} = 0$$
I am trying, but am unable to write ODE corresponding to this situation.
Best Answer
The equation of motion for constant acceleration is $$\frac{d^2x}{dt^2}=a$$ The distance covered is already given as $300$ meters. We still need to find the acceleration $a$. For that, let's rewrite the above equation as $$\frac{dv}{dt}=a$$ or $$dv=a\ dt$$ We integrate. At $t=0$ you have $v=v_0$, the initial velocity. After some time $t$ the velocity is $v$. $$\int_{v_0}^v dv=\int_0^t a\ dt\\v-v_0=a(t-0)\\v=v_0+at$$ Now we know that $v=\frac{dx}{dt}$, so we write the equation for $x$: $$\frac{dx}{dt}=v_0+at$$ Repeating the integration procedure, and assuming $x=0$ at $t=0$, you get $$\int_0^x dx=\int_0^t(v_0+at)dt\\x=v_0t+\frac12 at^2$$ You have now two unknowns, the acceleration of the car and the initial speed. But you can get this by knowing that the velocity after $10$ seconds is $0$, and the total distance traveled is $300$ meters. $$0=v_0+a\ 10\\300=v_0\ 10+\frac 12 a10^2$$ Solve these equations and you get your answer. Just a consistency check, the acceleration must be negative.