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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
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

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