Question

The velocity of a body changes from 65 m/s - 98 m/s by Time of 12 s . What is distance covered by body if acceleration is 2.75 m/s2.​

Answer 1

Given :

• The velocity of a body changes from 65 m/s to 98 m/s .
• Time taken by the body is 12 sec .
• Acceleration of the body is 2.75 m/s² .

To Find :

• Distance travelled by the body .

Solution :

$\longmapsto\tt{Initial\:Velocity\:(u)=65\:m/s}$

$\longmapsto\tt{Final\:Velocity\:(u)=98\:m/s}$

$\longmapsto\tt{Time\:Velocity\:(t)=12\:sec}$

Using 2nd Equation :

$\longmapsto\tt\boxed{s=ut+\dfrac{1}{2}\:a{t}^{2}}$

Putting Values :

$\longmapsto\tt{s=65\times{12}+\dfrac{1}{{\cancel{2}}}\times{2.75}\times{12}\times{{\cancel{12}}}}$

$\longmapsto\tt{s=780+2.75\times{72}}$

$\longmapsto\tt{s=780+198}$

$\purple\longmapsto\:\large\underline{\boxed{\bf\blue{s}\pink{=}\red{978\:m}}}$

So , The Distance travelled by the body is 978 m .

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Three Equations Of Motion :

• v = u + at
• s = ut + 1/2 at²
• v² - u² = 2as

Here :

• v = Final Velocity
• u = Initial Velocity
• t = Time Taken
• a = Acceleration
• s = Distance Travelled

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Answer 2

Answer:

$\underline{\purple{\ddot{MasterRohith}}}$

Given:-

• The velocity of the body changes from 65 m/s -98m/s by time of 12s.
• And also given acceleration is
• $= 2.75 \: m \: per \: {sec}^{2}$

To find :-

• Distance covered by the body.

Explanation :-

• Here we can find distance covered by the body by the 2nd equation of motion formula.

• Which is ,

• $s = ut + \frac{1}{2} a {t}^{2}$

• where,

• a=acceleration
• u=initial velocity
• t=time
• s=distance covered

• Now applying the values we get that,

• $s = 65 \times 12 + \frac{1}{2} \times 2.75 \times 12 \times 12$
• $s = 780 + 2.75 \times 72$
• $780 + 198$
• $s = 978m$

Therefore, the distance covered by the body is 978m.

☆Used formula :-

• Equations of motion .
• here we used 2nd equation of motion formula. which is s=ut+1/2at^2.
• So we get the distance covered by the body in this question.

Hope it helps u mate .

Thank you .