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

Answer:

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:

Distance covered by body if acceleration is 2.75 m/s² is 978 m.

Explanation:

Given that:

Initial velocity (u) = 65 m/s

Final velocity (v) = 98 m/s

Time taken (t) = 12 s

Acceleration (a) = 2.75 m/s²

To Find:

Distance covered (s)?

Solution:

Using second equation of motion ::

We know that,

⧪ \boxed{\bf{\purple{s = ut + \dfrac{1}{2}\:at^2}}}

s=ut+

2

1

at

2

⧪

Where,

s denotes distance covered

u denotes initial velocity

t denotes time taken

a denotes acceleration

According to the question by using the formula we get,

➼ \sf s = (65\:\times\:12) + \Bigg\{\dfrac{1}{2}\:\times\:2.75\:\times\:(12)^2\Bigg\}s=(65×12)+{

2

1

×2.75×(12)

2

}

➼ \sf s = 780 + \Bigg\{\dfrac{1}{\cancel{2}}\:\times\:2.75\:\times\:\cancel{144}\Bigg\}s=780+{

2

1

×2.75×

144

}

➼ \sf s = 780 + (2.75\:\times\:72)s=780+(2.75×72)

➼ \sf s = 780 + 198s=780+198

➼ \bf\pink{ s = 978\:m}s=978m

∴ Hence, distance covered by body if acceleration is 2.75 m/s² is 978 m.

Additional Information:

\clubsuit♣ Three equations of motion ::

v = u + at

s = ut + ½ at²ㅤㅤ[Used above]

v² = u² + 2as

\clubsuit♣ 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.

Answer 2

Answer:

initial velocity =65m/

Final velocity= 98m/s

time =12s

acceleration =2.75m/s²

$s = ut + 1 \2 {at}^{2}$

s represents distance covered

u represents initial velocity

t represents time

a represents acceleration

According to the formula:

s=(65×12)+{1/2×2.75×(12)²}

s=780+{1/2×2.75×144}

s=780+(2.75×72)

s=780+198

s=978 m

Therefore distance covered is 978m

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