Question

tke initial velocity of the particle is 10m/s and the retardation ithe initial velocity of a particle is 10m/sec and it's retardation is 2m/sec2.the distance covered in the 5th second of motion will be​

Answer 1

Answer:

1 metre

Explanation:

Given:

Initial velocity = u = 10 m/s

Acceleration = -2 m/s²

Time = 5 second

To find:

Distance covered in the 5th second of motion

The distance covered in nth second of motion is given by :

S (nth) = $u + \frac{a}{2} (2n-1)$

So substituting the above values, we get:

S (5th) = 10+$\frac{-2}{2} (2 \times 5 -1)$

S (5th) = 10+(-1(10-1))

S (5th) = 10+(-1(9))

S (5th) = 10-9

S (5th) = 1 metre

A distance of 1 metre is covered during the 5 th second of motion where the initial velocity is 10m/s and acceleration is equal to -2m/s²

Answer 2

Initial velocity = u = 10 m/s

Acceleration = -2 m/s²

Time = 5 second

To find:

Distance covered in the 5th second of motion

The distance covered in nth second of motion is given by :

S (nth) =

So substituting the above values, we get:

S (5th) = 10+

S (5th) = 10+(-1(10-1))

S (5th) = 10+(-1(9))

S (5th) = 10-9

S (5th) = 1 metre

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Explanation: