a body starting from rest travels a distance of 200m in 10 sec then the value of acceleration is
TPS:
s = 0.5 at^2
a = 4 m/s^2
Answers
Answered by
56
Initial velocity (u) = 0
distance (s) =200 m
Time (t) =10 sec
To find : Acceleration
S = ut + 1/2at²
200 = 0*10 + 1/2 (a*(10)²)
200= 1/2(a*100)
2*200 = a*100
400 = a*100
400/100 = a
4 = a
Or, a = 4 m/s²
distance (s) =200 m
Time (t) =10 sec
To find : Acceleration
S = ut + 1/2at²
200 = 0*10 + 1/2 (a*(10)²)
200= 1/2(a*100)
2*200 = a*100
400 = a*100
400/100 = a
4 = a
Or, a = 4 m/s²
if it helps you mark it brainliest
wrong answer
kaha se
answer ia 4
now ok
Answered by
28
hello friend....!!!
according to the question it is given ,
S = 200 m
t = 10 sec
u = 0 m/s
a = ???
we know ,
S = ut +at²
implies , 200 = ( 0 ) t + 1/2a(10)²
200 x 2 / 100 = a
therefore , a = 4 m/s²
the acceleration is 4 m/s² .
---------------------------------------------------------------
hope it helps...!!!
according to the question it is given ,
S = 200 m
t = 10 sec
u = 0 m/s
a = ???
we know ,
S = ut +at²
implies , 200 = ( 0 ) t + 1/2a(10)²
200 x 2 / 100 = a
therefore , a = 4 m/s²
the acceleration is 4 m/s² .
---------------------------------------------------------------
hope it helps...!!!
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