Question

A body, starting from rest and moving with a constant acceleration, covers 10 m in the first second. The distance travelled by it in the 2nd second is [ ]
a) 19 m
b) 10 m
c) 20m
d) 30 m​

Answer 1

Answer:

30 m [D]

Explanation:

As per the provided information in the given question, we have :

• Initial velocity, u = 0 m/s (Started from rest)
• Distance covered by it in 1st second, S′ = 10 m

We've been asked to calculate the distance travelled by it in the 2nd second.

As we know that, to find the distance travelled in nth second, we apply the formula given below :

⠀⠀⠀⠀⠀⠀⠀⠀$\underline{\boxed{ \textbf{\textsf{ S}} = \textbf{\textsf{u}} + \dfrac{ \textbf{\textsf{a}}}{ \textbf{\textsf{2}}} \textbf{\textsf{(2n - 1)}} }}$

According to the question, the distance covered by it in 1st second is 10 m. So,

$\\ \\ \longrightarrow\sf{S' = 0 + \dfrac{a}{2}\Bigg [2(1) - 1\Bigg ] }$

$\\ \\ \longrightarrow\sf{S' = 0 + \dfrac{a}{2}\Bigg [2- 1\Bigg ] }$

$\\ \\ \longrightarrow\sf{S' =\dfrac{a}{2}(1) }$

$\\ \\ \longrightarrow\sf{S' =\dfrac{a}{2} }$

$\\ \\ \longrightarrow\sf{2S' =a}$

$\\ \\ \longrightarrow\sf{2(10) =a}$

$\\ \\ \longrightarrow\underline{\underline{\sf{20 \; ms^{-2}=a}}}$

Now, let the distance covered by it in 2md second be S′′. Hence,

$\\ \\ \longrightarrow\sf{S'' = 0 + \dfrac{a}{2}(2n- 1)}$

$\\ \\ \longrightarrow\sf{S'' = \dfrac{20}{2}\Bigg [2(2) - 1\Bigg ] }$

$\\ \\ \longrightarrow\sf{S'' = 10\Bigg [4 - 1\Bigg ] }$

$\\ \\ \longrightarrow\sf{S'' = 10\Bigg [3\Bigg ] }$

$\\ \\ \longrightarrow\underline{\underline{\textbf{\textsf{S}}'' = \textbf{\textsf{30 \; m}} }}$

Therefore, the distance travelled by it in the 2nd second is 30 m.

⠀⠀⠀____________________________⠀⠀⠀

Answer 2

Answer:

Given :-

• A body starting from rest and moving with a constant acceleration, covers 10 m in the first second.

To Find :-

• What is the distance travelled by it in 2nd second.

Formula Used :-

$\clubsuit$ Distance Covered by nth second formula :

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

where,

• $\sf s_n$ = Distance Covered by nth second
• u = Initial Velocity
• a = Acceleration
• n = Number of second

Solution :-

First, we have to find the acceleration :

Given :

• Initial Velocity = 0 m/s
• Distance travelled in first second = 10 m

According to the question by using the formula we get,

$\implies \bf s_n =\: u + \dfrac{1}{2}a\bigg\{2n - 1\bigg\}$

where,

• n = 1

So, by putting this values we get,

$\implies \sf s_1 =\: 0 + \dfrac{1}{2} \times a\bigg\{2(1) - 1\bigg\}$

$\implies \sf (10)(1) =\: 0 + \dfrac{a}{2}\bigg\{2 \times 1 - 1\bigg\}$

$\implies \sf 10 \times 1 =\: 0 + \dfrac{a}{2}\bigg\{2 - 1\bigg\}$

$\implies \sf 10 - 0 =\: \dfrac{a}{2}\bigg\{1\bigg\}$

$\implies \sf 10 =\: \dfrac{a}{2} \times 1$

$\implies \sf 10 =\: \dfrac{a}{2}$

By doing cross multiplication we get,

$\implies \sf 10 \times 2 =\: a$

$\implies \sf 20 =\: a$

$\implies \sf\bold{\purple{a =\: 20\: m/s^2}}$

Now, we have to find the distance travelled by 2nd second :

Given :

• Initial Velocity = 0 m/s
• Acceleration = 20 m/s²

According to the question by using the formula we get,

$\dashrightarrow \bf s_n =\: u + \dfrac{1}{2}a\bigg\{2n - 1\bigg\}$

where,

• n = 2

So, by putting this values we get,

$\dashrightarrow \sf s_2 =\: 0 + \dfrac{1}{2} \times 20\bigg\{2(2) - 1\bigg\}$

$\dashrightarrow \sf s_2 =\: 0 + \dfrac{\cancel{20}}{\cancel{2}}\bigg\{2 \times 2 - 1\bigg\}$

$\dashrightarrow \sf s_2 =\: 0 + 10\{4 - 1\}$

$\dashrightarrow \sf s_2 =\: 10\{3\}$

$\dashrightarrow \sf s_2 =\: 10 \times 3$

$\dashrightarrow \sf\bold{\red{s_2 =\: 30\: m}}$

$\therefore$ The distance travelled by it in the 2nd second is 30 m .

Hence, the correct options is option no (d) 30 m .