A particle moves along x-axis. The acceleration of particle as function of time is given by a = 6t – 12 m/sec Initial velocity of particle is 9 m/sec and initially it was at origin, then find distance travelled by particle in first 2 seconds.
Answers
Answer:
Here we are given a = 6(t- 1) Here, s = +x, v = +v. To find velocity-time relation using a = dv/dt, we have
dt
dv
= 6 ( t - 1 )...(i)When velocity changes from v
0
to v during time t, integrating both sides with respect to time, we have∫
v
0
v
dv=∫
0
t
6(t−1)dt⇒v−v
0
=3t
2
−6t Substituting v
0
= 2 m/s, we have v =
dt
dx
=3t
2
−6t+2 (ii)dx = (3t
2
−6t+2)dt.(ii) find position-time relation, again integrating (ii) both sides, we have ∫
0
x
dx=∫
0
t
(3t
2
−6t+2)dt This gives x=t
3
−3t
2
+ 2t = t (t - 1)(t - 2) iii)The particle passes through the origin where x = 0 from (iii), we have t = 0, 1s and 2s. That means the particle crosses the origin twice at t = 1 s and t = 2 s. After t = 2 s, 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−
3
1
)s (=t
1
, say and t = (1+
3
1
)s (=t
2
, say)Hence, the particle at t =t
1
and t
2
. That signifies to and fro motion of the particle during first two seconds as it starts retracing its path at t = t
1
and t = t
2
The displacement (x), velocity (v), and acceleration (a) of the particle in different time intervals are given in the following table and shown in the following graphs.Time interval (1−
3
1
)(1−
3
1
) (1+
3
1
)
