What Are You Waiting For?
Sign In or Sign Up Now to begin your journey with us!
Invite your friends now. the more the merrier! Invite your friend to connect with each other.
:: Movie Contest Update ::
This contest has been closed.
Sign in or sign up
now to tell your friend about your referrer ID for use upon their registration. Cheers.
Weekly movie voucher contest is closed.
Maberlee weekly movie voucher contest is ended. Last pair of movie vouchers have been given out.
View past contest winners
A particle starts its motion from the origin and at time t its velocity parallel to coordinate axes remain (2t+3) and (4t) respectively.find the equation of its path.
August 26, 2016 at 3:51pm
If you have velocity, you have to integrate it to get displacement at different time, which also known as equation of its path. velocity vector = <2t+3, 4t> By integrating velocity, displacement vector = <2(t^2)/2 + 3t + c, 4(t^2)/2 + c> simplifying, you will get: displacement vector = <t^2 + 3t + c, 2t^2 + c> At t = 0, particle is at origin. So, x = 0, y = 0. By substituting t = 0, (0)^2 + 3(0) + c = 0 c = 0 2(0^2) + c = 0 c = 0 displacement or equation of its path = <t^2 + 3t, 2t^2>
August 27, 2016 at 1:25pm
Sign In or Sign Up
to reply to this post.
. All rights reserved. A Tian En Chong Forest production.