Question

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 ?

Answer 1

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

Answer 2

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

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