a body of mass 3kg moving with a constant acceleration covers a distance of 10m in the third second and lbm in the 4th second respectively.
.What is the initial velocity of the body ?
ShubhVidyarthi:
what is lbm
in the qns its like tht only !
but what does that mean
"lb" metres
I understood that thing but what is 'lb'
its not given !
the ans to the qn is -5 m/s
it must be 16 m
Where did u get distance covered in nth sec =u+a/2(2n-1)
Answers
Answered by
18
distance covered in nth second= u+a/2 (2n-1)
u= initial velocity
n= nth second
a= acceleration
distance covered in 3rd second= 10 m
therefore,
10= u+ a/2 (2*3-1)
10= (2u + 5a)/2
20= 2u+5a----eq.1
Similarly,
16= u+ a/2 (2*4-1)
16= (2u+7a)/2
32=2u + 7a----- eq.2
two equations, two variables
Solve it,
you'll get acceleration= 6m/s^2
and initial velocity= -5m/s
u= initial velocity
n= nth second
a= acceleration
distance covered in 3rd second= 10 m
therefore,
10= u+ a/2 (2*3-1)
10= (2u + 5a)/2
20= 2u+5a----eq.1
Similarly,
16= u+ a/2 (2*4-1)
16= (2u+7a)/2
32=2u + 7a----- eq.2
two equations, two variables
Solve it,
you'll get acceleration= 6m/s^2
and initial velocity= -5m/s
Answered by
13
Heya !
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in 3th second
S(3th) = 10 m
using
S(nth) = u + a/2(2n-1)
S(3rd) = u + a/2 (2x 3 -1)
10 = u + a/2 (5)
10 = u + 5a/2
MBS by 2
20 = 2u + 5a ------(1)
Similarly
S(4th) = u + a/2(2n-1)
16 = u + a/2(2 x 4 - 1)
16 = u + 7a/2
MBS by 2
32 = 2u + 7a ------(2)
By subtracting (1) from (2) we get ,
(32-20) = (2u-2u) + (7a - 5a)
12 = 0 + 2a
12 = 2a
a = 6 m/s²
Put value of a in 1st equation we get ,
20 = 2u + 5(6)
20 = 2u + 30
20 - 30 = 2u
-10 = 2u
-5 = u
hence u = -5 m/s
where negative (-ve) sign represents the direction of velocity
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in 3th second
S(3th) = 10 m
using
S(nth) = u + a/2(2n-1)
S(3rd) = u + a/2 (2x 3 -1)
10 = u + a/2 (5)
10 = u + 5a/2
MBS by 2
20 = 2u + 5a ------(1)
Similarly
S(4th) = u + a/2(2n-1)
16 = u + a/2(2 x 4 - 1)
16 = u + 7a/2
MBS by 2
32 = 2u + 7a ------(2)
By subtracting (1) from (2) we get ,
(32-20) = (2u-2u) + (7a - 5a)
12 = 0 + 2a
12 = 2a
a = 6 m/s²
Put value of a in 1st equation we get ,
20 = 2u + 5(6)
20 = 2u + 30
20 - 30 = 2u
-10 = 2u
-5 = u
hence u = -5 m/s
where negative (-ve) sign represents the direction of velocity
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