An object starts moving from rest on a straight line
path with uniform acceleration and covers a distance
of 100 m in first 10 s. The distance it covers in next
5 s is
Answers
Answered by
1
the distance covered in next 5 sec. is 25m.
nandini38:
25
25 m
please nandini explain me b/c i have forgotten how my sir explained me
hey the next 5 seconds mean that after 10 seconds so it is 15s
try to understand
if ur having any doubt refer other websites
you are telling total distance covered in above 10 soc. also
but the question says thatwhat is the distance covered in next 5 sec.
dude if you want this answer fir reference just take it
for*
Answered by
2
Now let us calculate the acceleration:
u = 0 m/s ; S = 100 m ; t = 10 s ; a = ?
S = ut^2 + 1/2 at^2
100 = 0 x 100 + 1/2 a x 100
100 = 50a
a = 2 m/s^2
Now let us take the time as 15 seconds because 10 s + 5 s and caculate the distance:
S = ? ; t = 15 s ; u = 0 m/s ; a = 2 m/s^2
S = ut^2 + 1/2 at^2
S = 0 x 225 + 1/2 x 2 x 225
S = 225 m
Therefore the distance covered in 15 s in 225 m.
u = 0 m/s ; S = 100 m ; t = 10 s ; a = ?
S = ut^2 + 1/2 at^2
100 = 0 x 100 + 1/2 a x 100
100 = 50a
a = 2 m/s^2
Now let us take the time as 15 seconds because 10 s + 5 s and caculate the distance:
S = ? ; t = 15 s ; u = 0 m/s ; a = 2 m/s^2
S = ut^2 + 1/2 at^2
S = 0 x 225 + 1/2 x 2 x 225
S = 225 m
Therefore the distance covered in 15 s in 225 m.
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