0

Guest

Courses New

at time t=0, a particle is at (2m, 4m). it starts moving towards positive x-axis with constant acceleration 2m/s^2 (initial velocity=0). after 2 s, an additional acceleration of 4m/s^2 starts acting on a particlein negative y-direction also. find after next 2 s.(1) velocity and (2) coordinates of particle.

at time t=0, a particle is at (2m, 4m). it starts moving towards positive x-axis with constant acceleration 2m/s^2 (initial velocity=0). after 2 s, an additional acceleration of 4m/s^2 starts acting on a particlein negative y-direction also. find after next 2 s.(1) velocity and (2) coordinates of particle.

Grade:11

3 Answers

Sitanshu kumar Singh
53 Points
5 years ago
First let`s find the velocity. First let`s write all the informational given in the question . Initial velocity =0,acceleration=2icap(as it moves towards positive x axis).and time =2s. then we can find the velocity using v=u+at and we get v=0+2icap×2=4icap.This is the velocity at the end of 2s.Now an additional acceleration starts acting so now a becomes 2icap-4jcap. So to find velocity after next 2 s we use v=u+at and it gives v=4icap(as calculated above)+(2icap-4jcap)×2= (8icap-8jcap). So now move further and calculate co-ordinate of the particle x=xnot +it+1/2at^2 and xnot is given as (2icap+4jcap) and u=0 and a =2icap and t=2s.so putting all these we get x=6icap+4jcap .this is the co-ordinate of particle after 2 s. Now to find the coordinates after next 2 s. we use the same equation and putting all values we get 18 icap-4jcap it means 18m,-4m Hope it helps u ask any queries relating to this if you didn`t understand please approved
Gitanjali Rout
184 Points
5 years ago
First let`s find the velocity. First let`s write all the informational given in the question . Initial velocity =0,acceleration=2icap(as it moves towards positive x axis).and time =2s. then we can find the velocity using v=u+at and we get v=0+2icap×2=4icap.This is the velocity at the end of 2s.Now an additional acceleration starts acting so now a becomes 2icap-4jcap. So to find velocity after next 2 s we use v=u+at and it gives v=4icap(as calculated above)+(2icap-4jcap)×2= (8icap-8jcap). So now move further and calculate co-ordinate of the particle x=xnot +it+1/2at^2 and xnot is given as (2icap+4jcap) and u=0 and a =2icap and t=2s.so putting all these we get x=6icap+4jcap .this is the co-ordinate of particle after 2 s. Now to find the coordinates after next 2 s. we use the same equation and putting all values we get 18 icap-4jcap it means 18m,-4m Hope it helps u ask any queries relating to this if you didn`t understand please approved
thank you
 
Eshan
askIITians Faculty 2095 Points
5 years ago
Due to constant accleration in the first 2 seconds, the coordinates of the particle becomes

(2+\dfrac{1}{2}at^2, 4)=(2+\dfrac{1}{2}(2)(2^2),4)=(6m,4m)

Due to additional deceleration in the y direction for the next 2 seconds, finally the coordinates of the partilcle becomes

(2+\dfrac{1}{2}(2)(4^2),4-\dfrac{1}{2}(4)(2^2))=(18m,-4m)

Think You Can Provide A Better Answer ?

Other Related Questions on Mechanics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More