• Support PF! Buy your school textbooks, materials and every day products Here!

Acceleration, velocity, and position.

  • Thread starter shark3189
  • Start date
  • #1
3
0

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)


Homework Equations


a(t) = v'(t) = d''(t)


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.
 

Answers and Replies

  • #2
Dr Transport
Science Advisor
Gold Member
2,273
413
There is an equation relating acceleration, velocity and distance......

hint, hint: it looks alot like conservation of energy......
 
  • #3
3
0
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.
 
  • #4
Dr Transport
Science Advisor
Gold Member
2,273
413
Try starting with integrating [tex] a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx} [/tex].
 
  • #5
3
0
ok, if [tex] a(t)=c_1 [/tex]

i get [tex] \int a(t)=v(t)=c_1 t+c_2[/tex]

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 [tex] v(0)=c_2=80.67 [/tex]

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

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

if i set that velocity equation equal to zero, will that give me the distance?
 
Last edited:
  • #6
Dr Transport
Science Advisor
Gold Member
2,273
413
How about integrating the third equality
[tex] a dx = v dv [/tex]

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

[tex] 2*a*s = v^{2} - v_{0} ^{2} [/tex]

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......
 

Related Threads for: Acceleration, velocity, and position.

Replies
3
Views
8K
  • Last Post
Replies
21
Views
583
  • Last Post
Replies
12
Views
3K
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
3K
Replies
1
Views
4K
Top