A car start from rest accelerate with 4 m/s Find 1) Velocity 2) Displacement Of t = 2 sec.
Answers
Explanation:
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.
Answer:
a car start from rest means u(initial velocity) =0 after 2 sec car velocity become V=u+at , u=0 , a = 4m/s^2 then V=8m/s , and S=ut+1/2at^2 where u= 0 so S=1/2at^2 S= 8m