Question

consider a toy car which is moving along a straight line initially at t=0 sec the car is at origin as shown in the figure.. First it moves towards west for 60 sec and at the end of 60 th second it reaches X=-10 m , there it stops for a movement and then starts moving along east for 100 seconds and at end of this time interval it is at X=+5 m . then it continues to move towards east and at the end of 250 th second it is at X=+20 m . calculate the average speed and average velocity​

Answer 1

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)+ 21.2( 210−t) 2 ]+v.tWhere

v=2.( 210−t )

Putting value of v in above equation we get=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.

So option C is the best possible answer.

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Answer 2

Answer:

Velocity formula,v = displacement/change in time.

While returning from West,it travelled -10m and returned to original position after 60seconds.

Then again it travelled 40 seconds and 150 seconds with total distance +5 and +20 m = 25m

Now,the velocity in West direction is ZERO ,because it reached the same spot.

Now,the velocity in East direction is :-

Distance = 25m and Time = 190 seconds.

Formula use :

v = d/t

v = 25/190

v = 0.13157m/s

Speed formula,s = distance /time

S = d/t

Let's see how much distance it travelled.

To the west :- -10m in 60seconds,

To the east :- 25m in 190seconds

Hence,total distance :- 25+10 = 35m

total time :- 60+40+150 = 250

Now,let's calculate the speed :-

S = d/t

S = 35/250

S = 0.14m/s

Sorry for any mistakes