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...
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.
Homework Statement
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)...