Question

the 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

HEYA!
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Given that

Initial velocity u = 10 m/s

Retardation = 2 m /s

Time = 5 sec .

Distance , s =?

Using second equation of motion,


$s = ut + \frac{1}{2} at {}^{2} \\ \\ s = 10 \times 5 + \frac{1}{2} \times - 2 \times (5) {}^{2}$


= 50 + ( -1 ) ( 25 )

= 50 -25


= 25m

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Answer 2

Answer:25m

Explanation:by second equation of motion s= u - a/2*2n-1

= 10-2/2*2*5-1 = 1m