A body starts from rest and has an acceleration of 20cm persecondsquare what is the distance covered by the body in first 8 seconds.
Answers
Answered by
13
Hi
______________________________________________________________
As We Know ,
S = ut+1/2 at ^2
Here ,
u = 0
a = 20
t = 8
S= ?
_________________________________________
S = 1/2 X 20 X 64
S = 640 cm
S = 6.4 m
_____________________________________________________________
______________________________________________________________
As We Know ,
S = ut+1/2 at ^2
Here ,
u = 0
a = 20
t = 8
S= ?
_________________________________________
S = 1/2 X 20 X 64
S = 640 cm
S = 6.4 m
_____________________________________________________________
kirangan6:
thanks sahillin
Answered by
8
HI there ...
As the body is starting from rest ,
Initial Velocity = u = 0 m/s
Acceleration = a = 20 cm/s²
Time = t = 8 sec
Distance traveled = s
Equation of Motion :
s = ut + 1/2at²
= 0 + 1/2 × 20 × 8 × 8
= 640 cm [ 1 m = 100 cm ]
= 640 /100 m
= 6.4 metre
As the body is starting from rest ,
Initial Velocity = u = 0 m/s
Acceleration = a = 20 cm/s²
Time = t = 8 sec
Distance traveled = s
Equation of Motion :
s = ut + 1/2at²
= 0 + 1/2 × 20 × 8 × 8
= 640 cm [ 1 m = 100 cm ]
= 640 /100 m
= 6.4 metre
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