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Asked by Prashant DIGHE | 22 Apr, 2020, 10:28: PM
Displacemet of particle,

x = 6 + 12t - 2t2 ......(1)

Q.6. time when velocity of particle is zero.

v = dx/dt
v = d (6 + 12 t - 2t2 )/dt

v = 12 - 4t ..... (2)

When v=0

0 = 12 - 4(t)
Thus, t = 12/4 = 3 s

Hence, when velocity becomes zero time is 3 s.

Q.7. Displacement of particle after 2 seconds.

Displacement after 2 s = s(t=2) - s (t = 0) = [6 + 12 (2) - 2 (2)2 ]  - [6] = 16 m

Q.8. Distance travelled after 2 seconds will be same as particle is moving in a straight line.
Thus, distance travelled = 16 m

Q.9.
Displacement of particle after 5 seconds.

Displacement after 5 s = s(t=5) - s (t = 0) = [6 + 12 (5) - 2 (5)2 ]  - [6] = 16 - 6 = 10 m

Thus, displacement after 5 sceonds is 10 m

Q.10.  Distance covered in first 5 s is,

In Q.6 we have seen that velocity becomes zero when t = 3s
It means particle travels forward direction for first 3 s and in backward direction for the next 2 s
Magnitude of displacement in first 3 seconds is
s(t=3) - s (t = 0) = [6 + 12 (3) - 2 (3)2 ]  - [6] = 24 -6 =18 m

then displacement during 3s to 5 s time interval is,
s(t=5) - s (t = 3) = [6 + 12 (5) - 2 (5)2 ]  - [6 + 12 (3) - 2 (3)2 ]  = 16 - 24 = -8  m

Total distance covered by the particle in first 5 seconds is = 18 + |-8| = 18 + 8 = 26 m
Answered by Shiwani Sawant | 23 Apr, 2020, 02:24: PM

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