please answer 7,8,9 and 10.

Asked by Prashant DIGHE | 22nd Apr, 2020, 10:28: PM

Expert Answer:

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 | 23rd Apr, 2020, 02:24: PM