A particle has an initial velocity of 9 m per second due east and a constant acceleration of 2 metre per second square the west the distance covered by the particle in the fifth second of its motion is
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hello friend...!!
• According to the question, we should calculate the distance covered in 5th second
• we know
Sn = u + a/2(2n-1)
it is given in the question that,
u = 9 m/s
a = 2 m/s^2
n = 5
therefore,
S5 = 9 + 2/2( 10-1)
S5 = 18 m
therefore the distance covered by the particle in 5th second is 18m
_____________________________
Hope it helps...!!
• According to the question, we should calculate the distance covered in 5th second
• we know
Sn = u + a/2(2n-1)
it is given in the question that,
u = 9 m/s
a = 2 m/s^2
n = 5
therefore,
S5 = 9 + 2/2( 10-1)
S5 = 18 m
therefore the distance covered by the particle in 5th second is 18m
_____________________________
Hope it helps...!!
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