A particle moves along the positive x-axis from origin with velocity, V0 = 2m/s and a = 6(t - 1) where't' is in seconds. Which of these graphs are correct?
A particle moves along the positive x-axis from origin with velocity, V0 = 2m/s and a = 6(t - 1) where't' is in seconds. Which of these graphs are correct?
The correct options are
A
C
Here we are given a = 6(t-1)
Here, s = +x, v = +v
To find velocity - time relation using a = dvdt we have dvdt=6(t−1) ........ (1)
When velocity changes from v0 to v during time t, integrating both sides with respect to time , we have
∫vv0dv=∫t06(t−1)dt⇒v−v0=3t2−6t
Substituting v0=2m/s, we have ,
v=dxdt=3t2−6t+2
dx = 3t2−6t+2) dt ..... (ii)
This gives x = t3−3t2+2t = t (t -1) (t - 2).
The particle passes through the origin where x = 0 from equation (iii), we have t = 0, 1s and 2s.
That means, the particle crosses the origin twice at t =1s and t = 2s. Aftert = 2s, x is positive. Hence its
displacement points in positive x-direction and goes on increasing its magnitude.
The particle will come at rest when v= 0.
From (ii), we have t = (1−1√3)s(=t1,say) and t=(1+1√3)s = (=t2,say).
Hence the particle at t = t1andt2. That signifies to and fro motion of the particle during first two seconds
as it starts retracing its path at time t = t1 and t=t2.
The displacement (x), velocity (v) and acceleration (a) of the particle in different time intervals are given
in the followingtable and shown in the following graphs.
A
C
Here we are given a = 6(t-1)
Here, s = +x, v = +v
To find velocity - time relation using a = dvdt we have dvdt=6(t−1) ........ (1)
When velocity changes from v0 to v during time t, integrating both sides with respect to time , we have
∫vv0dv=∫t06(t−1)dt⇒v−v0=3t2−6t
Substituting v0=2m/s, we have ,
v=dxdt=3t2−6t+2
dx = 3t2−6t+2) dt ..... (ii)
This gives x = t3−3t2+2t = t (t -1) (t - 2).
The particle passes through the origin where x = 0 from equation (iii), we have t = 0, 1s and 2s.
That means, the particle crosses the origin twice at t =1s and t = 2s. Aftert = 2s, x is positive. Hence its
displacement points in positive x-direction and goes on increasing its magnitude.
The particle will come at rest when v= 0.
From (ii), we have t = (1−1√3)s(=t1,say) and t=(1+1√3)s = (=t2,say).
Hence the particle at t = t1andt2. That signifies to and fro motion of the particle during first two seconds
as it starts retracing its path at time t = t1 and t=t2.
The displacement (x), velocity (v) and acceleration (a) of the particle in different time intervals are given
in the followingtable and shown in the following graphs.