Question

A particle of mass 0.5 kg is kept at rest. A force of 0.5 N acts on it for 5.0 s. Find

the distance moved by the particle in (a) these 5.0 s and (b) the next 5.0 s.​

Answer 1

Answer:

122.5m

Explanation:

formula : s=vit+1/2at²

s=?

vi=0

a =9.8

t=5s

s=(0)(5)+1/2(9.8)(5)²

s=0+(4.9)(25)

s=122.5 m

Answer 2

$\huge \mathfrak \pink{question}$

A particle of mass 0.5 kg is kept at rest. A force of 0.5 N acts on it for 5.0 s. Find

the distance moved by the particle in (a) these 5.0 s and (b) the next 5.0 s.

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a ) distance moved by participle in 5 second is 50 m

S = 0.5×4×10×10=200m

Then the distance moved in next 5 seconds is :

b ) 200-50 = 150

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