a particle moves 10m in 4th sec and 15m in 6th sec of its motion . calculate the distance travelled by it in 7th sec if the motion of the particle is uniformally accelerated ?
Answers
Answered by
60
Given : S4 = 10m ; S6 = 15m
Sn = u + a(n - 1/2)
⇒ u + a(4-1/2) = 10 ⇒ u+ 7a/2 = 10m ------ i
case ii = u + a(6 - 1/2) = 15 = u + 11a/2 = 15m ------ ii
On solving i and ii
we get a = 2.5 m/s² and u = 1.25m/s
Of S₇ = 1.25 + 2.5(7 - 0.5)
On simplification We Get S₇ = 17.5 m
Sn = u + a(n - 1/2)
⇒ u + a(4-1/2) = 10 ⇒ u+ 7a/2 = 10m ------ i
case ii = u + a(6 - 1/2) = 15 = u + 11a/2 = 15m ------ ii
On solving i and ii
we get a = 2.5 m/s² and u = 1.25m/s
Of S₇ = 1.25 + 2.5(7 - 0.5)
On simplification We Get S₇ = 17.5 m
Answered by
9
Answer:
Explanation:
Given :
S4 = 10m ; S6 = 15m
Sn = u + a(n - 1/2)
⇒ u + a(4-1/2) = 10 ⇒ u+ 7a/2 = 10m ------ i
case ii = u + a(6 - 1/2) = 15 = u + 11a/2 = 15m ------ ii
On solving i and ii
we get a = 2.5 m/s² and u = 1.25m/s
Of S₇ = 1.25 + 2.5(7 - 0.5)
On simplification We Get
S₇ = 17.5 m
Read more on Brainly.in - https://brainly.in/question/777317#readmore
Similar questions