Question

An object starts from rest and travels 7m with uniform acceleration in 4th second.. What will be the acceleration of the object?​

Answer 1

Answer:

The acceleration (a) of the object is 0.875 m/s²

Given:

1. Distance travelled (S) = 7 m

2. Time taken (t) = 4 sec.

3. Initial velocity (u) = 0 m/s

     ∵ [ Body starts from rest ]

Explanation:

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From the formula we know,

⇒ S = u t + 1 /2 a t²

Substituting the values,

⇒ 7 = 0 × 4 + 1 / 2 × a × (4)²

⇒ 7 = 0 + 1 / 2 × a × 16

⇒ 7 = 1 / 2 × a × 16

⇒ 7 × 2 = 16 a

⇒ 16 a = 14

⇒ a = 14 / 16

⇒ a = 0.875

⇒ a = 0.875 m/s²

∴ The acceleration (a) of the object is 0.875 m/s².

Note:

• Symbols have their usual meanings.

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Some Formulas:

⇒ S_n = u + a / 2 ( 2 n - 1 )

⇒ v² - u² = 2 a s

⇒ v = u + a t

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Answer 2

Given :

• Initial Speed (u) = 0 m/sec [As Object starts from rest ]

• Distance (s) = 7 metre

• Time (t) = 4 seconds

To Find :

• Acceleration Produced by the object

Solution :

Let's find acceleration from second equation of motion i.e.

$\bigstar \: \large \: \boxed{ \sf \: s = ut + \dfrac{1}{2} a {t}^{2} }$

Here,

• s = Distance covered

• u = Initial speed

• a = acceleration produced

• t = time taken

$\longrightarrow \sf \: 7 = (0)(4) + \dfrac{1}{2} (a)( {4})^{2}$

$\longrightarrow \sf \: 7 =( 0 \times 4) + \dfrac{1}{ \cancel2} \times a \times \cancel4 \times 4$

$\longrightarrow \sf \: 7 =0 +(2 \times 4 \times a)$

$\longrightarrow \sf \: 7 =0 + 8a$

$\longrightarrow \sf \: 7 = 8a$

$\longrightarrow \sf \: 8a = 7$

$\longrightarrow \sf \: a = \cancel\dfrac{7}{8}$

$\longrightarrow \sf \large \red{ a = 0.875m {s}^{ - 2} }$

$\therefore$ Acceleration produced by the object is 0.875m/s²