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 :

• Initial velocity of a body (u) = 65 m/s
• Final velocity of a body (v) = 98 m/s
• Time taken by body (t) = 12 seconds
• Acceleration of body (a) = 2.75 m/s²

To Find :

• Distance covered by body (s)?

Solution :

Using second eqⁿ of motion. We know that;

$\hookrightarrow\:{\large{\boxed{\sf{s = ut + \dfrac{1}{2}at^2}}}}$

$\spadesuit$ Putting all values :

$\longmapsto\:\tt s = (65\:\times\:12) + \dfrac{1}{2}\:\times\:2.75\:\times\:(12)^2$

$\longmapsto\:\tt s = 780 + \dfrac{1}{\cancel{2}}\:\times\:2.75\:\times\:\cancel{144}$

$\longmapsto\:\tt s = 780 + 2.75\:\times\:72$

$\longmapsto\:\tt s = 780 + 198$

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

Hence, distance covered by a body is 978 meters.

Learn More :

$\spadesuit$ Three equations of motion :

1. v = u + at
2. s = ut + ½ at²
3. v² = u² + 2as

Where,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time taken
• s denotes distance covered

$\spadesuit$ Some important definitions :

• Acceleration

Acceleration is the process where velocity changes. Since, velocity is the speed and it has some direction. So, change in velocity is considered as acceleration.

• Initial velocity

Initial velocity is the velocity of the object before the effect of acceleration.

• Final velocity

After the effect of acceleration, velocity of the object changes. The new velocity gained by the object is known as final velocity.

• Distance covered

The total path length covered by an object is known as distance covered.

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Answer 2

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$\large \bold \red{⚘\underline{ \underline{Given :-}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{Initial \: velocity \: (u) = 65 m/s}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{Final \: velocity \: (v) = 98 m/s}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{Time \: taken \: (t) = 12 \: seconds}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{Acceleration \: (a) = 2.75 m/s²}$

$\large \bold \red{⚘\underline{ \underline{To \: Find : - }}}$

$\bold{Distance }$

$\large \bold \red{⚘\underline{ \underline{Solution :-}}}$

$\bold{By \: Using \: 2 \: equation \: of \: motion }$

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$\pink➸\:{\large{ \underline{\tt{s = ut + \dfrac{1}{2}at^2}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red➭\:\bold s = 65\:\times\:12 + \dfrac{1}{2}\:\times\:2.75\:\times\:(12)^2$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red➭\:\bold s = 780 + \dfrac{1}{\cancel{2}}\:\times\:2.75\:\times\:\cancel{144}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\red➭\:\bold s = 780 + 2.75\times\:72$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red➭\:\bold s = 780 + 198$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dashrightarrow \underline{ \boxed{ \frak \pink{s = 980 \: m}}} \bigstar$

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Concept Used here:-

$\pink{\large \qquad \boxed{\boxed{\begin{array}{cc} \bigstar \: \: \bf v = u + at \\ \\ \bigstar\boxed{ \bf s = ut + \dfrac{1}{2}a {t}^{2} }\\ \\ \bigstar\: \: \bf{v}^{2} - {u}^{2} = 2as\end{array}}}}$

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