Acceleration, velocity, and position.

[shark3189]

1. The problem statement, all variables and given/known data
The speed of a car traveling in a straight line is reduced from 55 to 40 mph in a distance of 362 feet. Find the distance in which the car can be brought to rest from 40 mph, assuming the same constant deceleration. (You must show the calculus behind your response)


2. Relevant equations
a(t) = v'(t) = d''(t)


3. The attempt at a solution
Since acceleration is constant,
a(t) = c1
v(t) = c1*x + c2
d(t) = .5*c1*x^2 + c2*x + c3
I have no idea where to go from here.

[Dr Transport]

There is an equation relating acceleration, velocity and distance......

hint, hint: it looks alot like conservation of energy......

[shark3189]

yea but my teacher said any equations we use we have to derive, starting from acceleration. from there i integrated the others but i don't know what to plug in.

[Dr Transport]

Try starting with integrating a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx} .

[shark3189]

ok, if a(t)=c_1

i get \int a(t)=v(t)=c_1 t+c_2

then i converted everything to feet and seconds, so the velocities are 80.67 ft/sec and 58.67 ft/sec.

i plugged in t=0 and my equation was v(0)=c_2=80.67

i also got c_1=\frac{80.67-58.67}{0-362}=-\frac{11}{181}

so the velocity equation is v(t)=\frac{-11t}{181}+80.67

if i set that velocity equation equal to zero, will that give me the distance?

[Dr Transport]

How about integrating the third equality
a dx = v dv

\int_{0} ^{s} a dx = \int_{v_{0}} ^{v} v dv , a is a constant

2*a*s = v^{2} - v_{0} ^{2}

you have the distance and the initial and final velocities, calculate the acceleration and the plug into the very same equation to find how far it will take to stop the car......