Question

A body suffers a displacement of 9m in 3rd second and 15m in 6th second of motion . The velocity acquired by it at the end of 5th second is​

Answer 1

Given :

• Displacement in 3rd second = 9m
• Displacement in 6th second = 15m

Aim :

• To find the displacement in 5th second.

Solution :

Using the distance travelled in nth second formula,

$\boxed{ \longrightarrow \sf S_n = u + \dfrac{1}{2} a(2n - 1)}$

Here,

• S = Displacement travelled in nth second
• u = Initial velocity
• a = Acceleration
• n = Time

According to the question,

for displacement in 3rd second,

$\implies \sf 9 = u + \dfrac{1}{2} a(2 \times 3 - 1)$

$\implies \sf 9 = u + \dfrac{1}{2} a(5)$

$\implies \sf 9 = u + \dfrac{5a}{2}$

for displacement in 6th second,

$\implies \sf 15 = u + \dfrac{1}{2}a(2 \times 6 - 1)$

$\implies \sf 15 = u + \dfrac{11a}{2}$

By using elimination method,

(EQUATION 2) - (EQUATION 1)

$\implies \sf \bigg(15 = u + \dfrac{11a}{2} \bigg) - \bigg(9 = u + \dfrac{5a}{2} \bigg)$

$\implies \sf 6 = \dfrac{6a}{2}$

Reducing to the lowest terms,

$\implies \sf 6 = 3a$

$\implies \sf \dfrac{6}{3} = a$

$\implies \sf 2 = a$

Hence, from the above solutions, acceleration = 2m/s².

Substituting this value in one of the Equations, we get u = 4m/s

Velocity at the end of 5 seconds :

→ Using the first Equation of motion, v = u + at, let us find the final velocity.

→ v = 4 + 2(5)

→ v = 4 + 10

→ v = 14m/s

Therefore, the final velocity after 5 seconds is 14m/s

More formulas :

• v = u + at
• s = ut + ½at²
• v² = u² + 2as

$\rightarrow \sf speed = \dfrac{distance}{time}$

$\rightarrow \sf velocity = \dfrac{displacement}{time}$

$\rightarrow \sf acceleration = \dfrac{velocity}{time}$

Answer 2

Answer:

Given :

Displacement in 3rd second = 9m

Displacement in 6th second = 15m

Aim :

To find the displacement in 5th second.

Solution :

Using the distance travelled in nth second formula,

\boxed{ \longrightarrow \sf S_n = u + \dfrac{1}{2} a(2n - 1)}⟶Sn=u+21a(2n−1)

Here,

S = Displacement travelled in nth second

u = Initial velocity

a = Acceleration

n = Time

According to the question,

for displacement in 3rd second,

\implies \sf 9 = u + \dfrac{1}{2} a(2 \times 3 - 1)⟹9=u+21a(2×3−1)

\implies \sf 9 = u + \dfrac{1}{2} a(5)⟹9=u+21a(5)

\implies \sf 9 = u + \dfrac{5a}{2}⟹9=u+25a

for displacement in 6th second,

\implies \sf 15 = u + \dfrac{1}{2}a(2 \times 6 - 1)⟹15=u+21a(2×6−1)

\implies \sf 15 = u + \dfrac{11a}{2}⟹15=u+211a

By using elimination method,

(EQUATION 2) - (EQUATION 1)

\implies \sf \bigg(15 = u + \dfrac{11a}{2} \bigg) - \bigg(9 = u + \dfrac{5a}{2} \bigg)⟹(15=u+211a)−(9=u+25a)

\implies \sf 6 = \dfrac{6a}{2}⟹6=26a

Reducing to the lowest terms,

\implies \sf 6 = 3a⟹6=3a

\implies \sf \dfrac{6}{3} = a⟹36=a

\implies \sf 2 = a⟹2=a

Hence, from the above solutions, acceleration = 2m/s².

Substituting this value in one of the Equations, we get u = 4m/s

Velocity at the end of 5 seconds :

→ Using the first Equation of motion, v = u + at, let us find the final velocity.

→ v = 4 + 2(5)

→ v = 4 + 10

→ v = 14m/s

Therefore, the final velocity after 5 seconds is 14m/s

HOPE IT HELPS!!!!!!!