Question

A body travels 59m in the fifth second and 69m in the tenth second. Its acceleration is ____________m/s2 and its initial velocity is _______m/s.

Answer 1

Answer :-

→ Acceleration = 2 m/s²

→ Initial velocity = 50 m/s

Explanation :-

We have :-

→ Distance travelled in 5th sec = 59 m

→ Distance travelled in 10th sec = 69 m

______________________________

For the 5th second :-

⇒ sₙₜₕ = u + a/2(2n - 1)

⇒ 59 = u + a/2(2 × 5 - 1)

⇒ 59 = u + 9a/2

⇒ 59 = u + 4.5a ----(1)

For the 10th second :-

⇒ sₙₜₕ = u + a/2(2n - 1)

⇒ 69 = u + a/2(2 × 10 - 1)

⇒ 9 = u + 19a/2

⇒ 69 = u + 9.5a ----(2)

______________________________

Subtracting eq.1 from eq.2, we get :-

⇒ 69 - 59 = u + 9.5a - (u + 4.5a)

⇒ 10 = u + 9.5a - u - 4.5a

⇒ 10 = 5a

⇒ a = 10/5

⇒ a = 2 m/s²

Now, putting value of 'a' in eq.1 :-

⇒ 59 = u + 4.5(2)

⇒ 59 = u + 9

⇒ u = 59 - 9

⇒ u = 50 m/s

Answer 2

Answer:

Given :-

• A body travels 59 m in the fifth second and 69 m in the tenth second.

To Find :-

• What is the acceleration and the initial velocity of a body.

Solution :-

❒ In case of fifth second :

As we know that :

$\bigstar\: \: \sf\boxed{\bold{\pink{s_{nth} =\: u + \dfrac{1}{2}a\bigg\{2n - 1\bigg\}}}}\: \: \bigstar$

where,

• s = Distance Covered
• u = Initial Velocity
• a = Acceleration
• n = Time Period in seconds

Given :

• Distance Covered $\sf (s_{nth})$ = 59 m
• Time Period (n) = 5 seconds

According to the question by using the formula we get,

$\implies \sf 59 =\: u + \dfrac{1}{2}a\bigg\{2(5) - 1\bigg\}$

$\implies \sf 59 =\: u + \dfrac{a}{2}\bigg\{2 \times 5 - 1\bigg\}$

$\implies \sf 59 =\: u + \dfrac{a}{2}\bigg\{10 - 1\bigg\}$

$\implies \sf 59 =\: u + \dfrac{a}{2}\bigg\{9\bigg\}$

$\implies \sf 59 =\: u + \dfrac{9a}{2}$

$\implies \sf\bold{\purple{59 =\: u + 4.5a\: ------\: (Equation\: No\: 1)}}$

❒ In case of tenth second :

Given :

• Distance Covered $\sf (s_{nth})$ = 69 m
• Time Period (n) = 10 seconds

According to the question by using the formula we get,

$\implies \sf 69 =\: u + \dfrac{1}{2}a\bigg\{2(10) - 1\bigg\}$

$\implies \sf 69 =\: u + \dfrac{a}{2}\bigg\{2 \times 10 - 1\bigg\}$

$\implies \sf 69 =\: u + \dfrac{a}{2}\bigg\{20 - 1\bigg\}$

$\implies \sf 69 =\: u + \dfrac{a}{2}\bigg\{19\bigg\}$

$\implies \sf 69 =\: u + \dfrac{19a}{2}$

$\implies \sf\bold{\purple{69 =\: u + 9.5a\: ------\: (Equation\: No\: 2)}}$

From the equation no 1 and 2 we get,

$\implies \sf u + 4.5a - (u + 9.5a) =\: 59 - (69)$

$\implies \sf u + 4.5a - u - 9.5a =\: 59 - 69$

$\implies \sf {\cancel{u}} {\cancel{- u}} + 4.5a - 9.5a =\: - 10$

$\implies \sf 4.5a - 9.5a =\: - 10$

$\implies \sf {\cancel{-}} 5a =\: {\cancel{-}} 10$

$\implies \sf 5a =\: 10$

$\implies \sf a =\: \dfrac{\cancel{10}}{\cancel{5}}$

$\implies \sf\bold{\red{a =\: 2\: m/s^2}}$

Now, again by putting a = 2 in the equation no 2 we get,

$\implies \sf 69 =\: u + 9.5a$

$\implies \sf 69 =\: u + 9.5(2)$

$\implies \sf 69 =\: u + 9.5 \times 2$

$\implies \sf 69 =\: u + 19$

$\implies \sf 69 - 19 =\: u$

$\implies \sf 50 =\: u$

$\implies \sf\bold{\red{u =\: 50\: m/s}}$

${\footnotesize{\bold{\underline{\therefore\: The\: acceleration\: is\: 2\: m/s^2\: and\: the\: initial\: velocity\: is\: 50\: m/s\: .}}}}$