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 | 14th May, 2021, 11:18: AM

Expert Answer:

### Velocity at end of 4^{th} second , v_{4} = v_{o} - ( a t ) = 9 - ( 2 × 4 ) = 1 m/s
Displacement at 5^{th} second , S = ( u t ) - [ (1/2) a t^{2} ] = ( 1 ×1 ) - ( 0.5 × 2 × 1 × 1 ) = 0
Above expression shows the displacement is zero .
Hence just after 5^{th }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 5^{th} second is given as,
t = u/a = initial speed / acceleration = 1 / 2 = 0.5 s
Distance moved in 0.5 s after 5^{th} second , S = ( ut ) -(1/2) at^{2} = ( 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

^{th}second , v

_{4}= v

_{o}- ( a t ) = 9 - ( 2 × 4 ) = 1 m/s

^{th}second , S = ( u t ) - [ (1/2) a t

^{2}] = ( 1 ×1 ) - ( 0.5 × 2 × 1 × 1 ) = 0

^{th }second particle moves forward with velocity 1 m/s then it is stopped then it turns back

^{th}second is given as,

^{th}second , S = ( ut ) -(1/2) at

^{2}= ( 1 × 0.5 ) - ( 0.5 × 2 × 0.5 × 0.5 ) = 0.25 m

### Answered by Thiyagarajan K | 14th May, 2021, 12:41: PM

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