A particle of mass 0.5kg is kept at rest.A force of 2.0 N act on it for 5 seconds.Find the distance moved by the particlein
A)first 5 seconds
B)Next 5 seconds
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a= F/m
a=2/0.5=4m/s²
Distance covered in first 5 sec
s=ut +1/2at²
s=1/2*4*5²
s=50m
Now we will find distance covered in first 10 sec
s=1/2*4*10²
s=200m
Therefore, the distance travelled in next 5 sec is 200-50=150m
a=2/0.5=4m/s²
Distance covered in first 5 sec
s=ut +1/2at²
s=1/2*4*5²
s=50m
Now we will find distance covered in first 10 sec
s=1/2*4*10²
s=200m
Therefore, the distance travelled in next 5 sec is 200-50=150m
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