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# solve itt

Asked by mehtaswati566 30th December 2022, 10:36 AM
Let S4 be the distance covered in 4th second

Let us use the equation of motion , " S = u t + (1/2) a t2 " to find the distance S travellled
in t = 1 s if acceleration a is 4 m/s2 and u is initial velocity .

S4 = u3  + (1/2) (4) = 20 m

where u3 is velocty after travelling 3 seconds.

we get u3 = ( 20 - 2 ) m/s = 18 m/s

If uo is initial velocity at t=0 s , then we have

u3 = uo + a t  = uo + ( 4 × 3 ) = 18

Hence , initial velocity uo = ( 18 - 12 ) = 6 m/s

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Let S3 be the distance covered in 3rd second

S3 = u2 + (1/2)a

where u2 is velocity after 2 seconds

S3 = ( uo + 2 a ) + (1/2) a = uo + (5/2) a = 6 + ( 2.5 × 4 ) = 16 m/s

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Let S5 be the distance covered in 5th second

S5 = u4 + (1/2)a

where u4 is velocity after 4 seconds

S5 = ( uo + 4 a ) + (1/2) a = uo + (9/2) a = 6 + ( 4.5 × 4 ) = 24 m/s

Answered by Expert 31st December 2022, 12:35 PM
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