A car is traveling at $50 mi /h$ when the brakes are fully applied, producing a constant deceleration of $38 ft/s^2$. What is the distance covered before the car comes to a stop?
Since the car is decelerating, our value will be negative.
$v'(t)= a(t)=-38$ so $v(t)= -38x= s(t)-38x^2/(2)$ After this point I am not sure what I am supposed to do? Help would be appreciated.
Best Answer
Well let us first attempt to get our units the same. For miles/hour we can rewrite this as $\frac{220}{3}$ feet per second.
We are given an acceleration, an initial velocity, and are asked for a distance.
So, $$\frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = a \implies vdv = adx$$
Integrating from initial velocity given to 0, and using the given acceleration, we find:
$$0 - \frac{1}{2}(\frac{220}{3})^2 = (-38) \Delta x$$
Or,
$$\Delta x = 70.7602 ft$$