An object starts from rest and travels 7m with uniform acceleration in 4th second.. What will be the acceleration of the object?
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An object starts from rest and travels 7m with uniform acceleration in 4th second.. What will be the acceleration of the object?
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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:
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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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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.
[tex] \bigstar \: \large \: \boxed{ \sf \: s = ut + \dfrac{1}{2} a {t}^{2} }[/tex]
Here,
• s = Distance covered
• u = Initial speed
• a = acceleration produced
• t = time taken
[tex] \longrightarrow \sf \: 7 = (0)(4) + \dfrac{1}{2} (a)( {4})^{2} [/tex]
[tex] \longrightarrow \sf \: 7 =( 0 \times 4) + \dfrac{1}{ \cancel2} \times a \times \cancel4 \times 4[/tex]
[tex] \longrightarrow \sf \: 7 =0 +(2 \times 4 \times a)[/tex]
[tex] \longrightarrow \sf \: 7 =0 + 8a[/tex]
[tex] \longrightarrow \sf \: 7 = 8a[/tex]
[tex] \longrightarrow \sf \: 8a = 7[/tex]
[tex] \longrightarrow \sf \: a = \cancel\dfrac{7}{8} [/tex]
[tex] \longrightarrow \sf \large \red{ a = 0.875m {s}^{ - 2} }[/tex]
[tex]\therefore[/tex] Acceleration produced by the object is 0.875m/s²