Breaks are applied to a car to produce an acceleration of 6ms- in the opposite direction to the motion. If the car takes 2seconds to stop after the application of breaks. Calculate the distance travelled during this time.
Answers
Answered by
1
HEY MATE
AS .
ACCELERATION =6m/s2
time=2sec
as
S=UT+1/2AT²
AS INITIAL VELOCITY IS ZERO THEN
S=1/2AT²
S=1/2×6×2²
S=12meter
SO DISTANCE COVERED BY THE PARTICLE IS 12 METER
HOPE HELP U
AS .
ACCELERATION =6m/s2
time=2sec
as
S=UT+1/2AT²
AS INITIAL VELOCITY IS ZERO THEN
S=1/2AT²
S=1/2×6×2²
S=12meter
SO DISTANCE COVERED BY THE PARTICLE IS 12 METER
HOPE HELP U
Answered by
3
I think it's ans. is -12m/s2 as decelaration.
(Not sure but,I think)
As ATQ,
u=0
a=-6m/s2
t=2 sec.
s=ut+1/2at2
s=0×2+1/2×(-6)×4
=0+(-12)
=-12 ( I think.)
(Not sure but,I think)
As ATQ,
u=0
a=-6m/s2
t=2 sec.
s=ut+1/2at2
s=0×2+1/2×(-6)×4
=0+(-12)
=-12 ( I think.)
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