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

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Answer 1

Answer:

The distance covered by the body is 978 m.

Explanation:

As the initial velocity of the body is,

$u = 65 \: m {s}^{ - 1}$

The final velocity of the body is,

$v = 98 \: m {s}^{ - 1}$

The acceleration of the body is,

$a = 2.72 \: m {s}^{ - 2}$

By using the kinematics equation, the value of the distance covered by the body is,

${v}^{2} - {u}^{2} = 2ad \\ {98}^{2} - {65}^{2} = 2 \times 2.75 \times d \\ 9604 - 4225 = 5.5 \times d \\ 5379 = 5.5 \times d \\ d = \frac{5379}{5.5} \\ d = 978 \: m$

9

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

Answer 2

Answer:

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