1yes ketan i agree with u
u can also try using diffrentiation
dv/dt=a=6t-6
now put t=3.5
u will get avg acceleration
now just use simple eqn. s=ut+(1/2)at2
now u will get ur answer
still i maybe wrong
A particle moves along a horizontal path,such that its velocity is given by v=(3t2-6t)m/s,where t is the time in seconds.If it is initially located at the origin O,determine the distance travelled by the particle in time interval from t=0 to t=3.5s.
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15 Answers
1057integrate (3t2-6t)dt from 0 to 3.5 secs....you will get the answer....
11057there is a twist here....we have to find the distance and not the displacement....
hope you get a hint from it....
1but also note tat there iz nt mentioned anything about the motion of the particle so considering it to be linear motion distance travelled=displacement......(in case of linear motion)
1057nopes.....
distance≥displacement.....
they are not necessarily equal....
1i said in linear motion
distance=displacement
in all other cases distance≥diplacement
1yes it is true try take an example and just think about it that a train is travelling in a straight line then velocity=speed
=>displacement/time=distance/time
=>displacement=distance
1057arre....cant the engine of the train go mad and it started moving in the opposite direction and reached the starting point....in this case displacement will be zero and distances two times the farthest point the train reached....got it?
49@Ketan: even in ur method you are finding out the displacement and not the distance traversed.
49@Rest: See, the particle is changing the direction of motion at t=2s.
After this time, the body's direction of motion (the sign of the velocity changes)
Thus, the net distance traversed = magnitude of displacement from t=0s to t=2s + magnitude of displacement from t=2s to t=3.5s
1057@subho da.....i know my first post is wrong....that's why i wrote about the twist....
158see if u get displacement from t=0 to 3.5 s,u get 6.125 m
and if u get displacement from t=0s to 2s , u get -4m
and displacement from t=2s to 3.5 s ,u get 10.125m
so ...the answr is the same...!!! given correct answr is 14.125m
1057arre bhai.....distances is sum of the magnitudes...direction doesnt matter and so does the negative sign....so your answer will be 10.125+4=14.125m....
got it?