Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Introductory Physics Homework Help
Newton's Second Law Problem
Reply to thread
Message
[QUOTE="Rawrr!, post: 4531420, member: 490024"] In 32a) to find the distance, you seemed to have used the mass of the car, in place of its velocity. That being said, i think your equation could be modified. The question is relating to the distance between the car and truck while in motion. the velocity remains constant. distance = (velocity)x(time), *make sure to convert the components into correct units. (km/h) -> (km/s) (since time is given in seconds)* *you will get a distance in terms of (km) when you do this, which you can then convert to (m) if you feel so compelled* in b) When dealing with forces, it always helps to draw a force body diagram, letting you visualize the situation and see which direction the forces are going. This always helps me in determining the correct signs as well. You have a force in the left direction, (by convenience, placing the car's velocity along the positive x-axis), meaning your stopping force is going to be negative. -F[SUB]net[/SUB] = m[SUB]car[/SUB] x (-)a This should make sense as the car is slowing down, the force is in the negative direction, and the car is accelerating to the left. While you have the correct signs, you seemed to have neglected to convert the (km/h) of the velocity, into the (km/s) needed, since your time component is in seconds. It always helps to think about you answers in terms of the situation. With the time you have written down in your answer, it means it will take the car 226.4 seconds to come to a stop. (Which means it will take you over 3 minutes to come to a stop!) So i would always beg you to make sure you answer makes sense to the situation. in 3c) you seemed to have done the kinematic equations correctly in this and the equations make sense in that the driver will travel a distance during reacting, and will also travel a distance during the braking, but again avoided to convert the (km/h) of the velocity, into (km/s). In order for the units to cancel, the time parameters must match. *Write down your units as well as the value into the equations, to make sure they cancel correctly, and that you finish with what you need! :tongue: * I hope this helps!p.s. - Welcome to PF! [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
Newton's Second Law Problem
Back
Top