Request a call back

Join NOW to get access to exclusive study material for best results

CBSE Class 11-science Answered

a particle has initial velocity of 9 m/s^2 and retardation of 2m/s^2.find the distance travelled in 5 th second

Asked by trenuka180 | 14 May, 2021, 11:18: AM
Expert Answer
Velocity at end of 4th second , v4 = vo - ( a t ) = 9 - ( 2 × 4 ) = 1 m/s
 
Displacement at 5th second , S = ( u t ) - [ (1/2) a t2 ]  = ( 1 ×1 ) - ( 0.5 × 2 × 1 × 1 ) = 0
 
Above expression shows the displacement is zero .
 
Hence just after 5th second particle moves forward with velocity 1 m/s then it is stopped then it turns back
in opposite direction if retardation continues.
 
Time taken for the particle to reach the zero speed after 5th second is given as,
 
t = u/a  = initial speed / acceleration = 1 / 2 = 0.5 s
 
Distance moved in 0.5 s after 5th second , S = ( ut ) -(1/2) at2  = ( 1 × 0.5 ) - ( 0.5 × 2 × 0.5 × 0.5 ) = 0.25 m
 
Hence particle has moved 0.25 m forward and then 0.25 m backward
 
Total distance travelled = ( 0.25 + 0.25 ) m = 0.5 m
 

Answered by Thiyagarajan K | 14 May, 2021, 12:41: PM
CBSE 11-science - Physics
Asked by ukalyadab | 18 Jul, 2022, 11:52: AM
ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by piyushhoney2006 | 20 Jun, 2022, 04:12: PM
ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by kalraneha629 | 18 Jun, 2022, 09:16: PM
ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by chinttubabu9692 | 27 Jan, 2022, 06:37: PM
ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by anyone.viv7807191905 | 21 Dec, 2021, 11:40: PM
ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by happy2001abhishek | 05 Nov, 2021, 09:49: PM
ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by erkyperky2004 | 03 Nov, 2021, 12:23: PM
ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by abhishekjangid3102002 | 15 Oct, 2021, 06:00: AM
ANSWERED BY EXPERT
×