A car moves with
moves with uniform
velocity of 40 ms for 5 s. It
comes to rest in the next 10 s
with uniform deceleration. Find
() deceleration (ii) total
distance travelled by the car.
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Answer:
ANSWER
As average speed is given so distance or displacement of the body is given by 3.2×10=32m
Distance covered while acceleration and retardation are the same. because time taking to 0 to v is equal to that of v to 0 as the value of acceleration and retardation is the same. Using v
2
−u
2
=2as
Break the journey in 3 parts acceleration, retardation, and uniform motion.
And let uniform motion is performed for t sec.
So 32=2[0.
2
(10−t)
+
2
1
.2(
2
10−t
)
2
]+v.t
Where v=2.(
2
10−t
)
Putting value of v in above equation we get t=6sec
So the body will move with uniform velocity for middle 2 sec and accelerated for first w sec and retarded for the last 2 sec.
HOPE IT HELPS YOU
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