please solve the question in the figure
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the initial velocity of the car at a= 160km/hr
the final velocity of the car at c= 40km/hr
the let the distance between the equal intervals be x
therefore the distance covered=2x
so ,v²=u²+2as
40²=160²+2×2x×a
1600-25600=+4ax
-24000=+4ax
ax=-6000
so the distance covered when the final velocity is 100km/hr
(100)²=(160)²+2aS
10000-25600=2aS
15600=-2aS
-7800=aS
as the acceleration is uniform through out and acceleration is negative we assume it to be -a
so x=6000 but S=7800
car is between mark b and c because distance between a and c would be 12000 so it must be in between
MARK ME AS BRAIN LIST
the final velocity of the car at c= 40km/hr
the let the distance between the equal intervals be x
therefore the distance covered=2x
so ,v²=u²+2as
40²=160²+2×2x×a
1600-25600=+4ax
-24000=+4ax
ax=-6000
so the distance covered when the final velocity is 100km/hr
(100)²=(160)²+2aS
10000-25600=2aS
15600=-2aS
-7800=aS
as the acceleration is uniform through out and acceleration is negative we assume it to be -a
so x=6000 but S=7800
car is between mark b and c because distance between a and c would be 12000 so it must be in between
MARK ME AS BRAIN LIST
sanidhyasingla:
if acceleration is uniform how can you cut' a' out
and say that x=6000
because whatever the acceleration will be
we will have cancel the same in the equation
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