Question

Using Integrals. A car traveling at 45 miles per hour is brought to a stop, at constant deceleration, 184 feet from where the brakes are applied.
1) How far has the car moved when its speed has been reduced to 30 miles per hour? (Answer in feet)
2) How far has the car moved when its speed has been reduced to 15 miles per hour? (Answer in feet)

Answer 1

Please help me!

Answer 2

Since acceleration is constant we can use the kinematic equations but use integral calculus to derive them.

a = acceleration and its constant \[a=\frac{ dv }{ dt }\]
We to solve for dv then integrate.
\[dv = adt\]
\[v _{f}-v _{i} = \int\limits_{ti}^{tf} adt\]
You get after integration.
\[v _{f}= v _{i} + a (t _{f}-t _{i})\] 1.1

This will give you velocity but we want to know position so we have to integrate again.
\[r _{f} -v _{i}= \int\limits_{ti}^{tf}(v _{i}+at)dt\]

After integration you get\[r _{f}=r _{i}+v _{i}t+at ^{2}\]

Since we have deceleration the acceleration will be negative. so put a negative sign in front of a.

Answer 3

Starting with an acceleration "a" you can derive the velocity and distance functions using integration
\[v(t) =a \int\limits dt = at + v_o\]
\[d(t) = \int\limits v(t) dt = \frac{a}{2} t^2 +v_o t + d_o\]
initial distance is 0,

2 unknowns , time and acceleration

use velocity equation to solve for "t"
\[at + 45 = 0\]
\[t = -\frac{45}{a}\]
substitute into distance equation
\[\frac{a}{2} (-\frac{45}{a})^2 + 45(\frac{-45}{a}) =\frac{184}{5280}\]
\[a = (-\frac{45^2}{2})(\frac{5280}{184})\]

with the deacceleration now known, you can plug in new velocity and solve for distance using distance equation