Question

А A particle accelerates from rest
constant rate for Some time and attains
constant velocity of Pms-4. After words it decedere
constant rate
and comes
to
If the total time taken is 4 seconds ,the
distance traveled is.
with a​

Answer 1

Explanation:

Using v=u+at

During acceleration,8=at

1

ort

1

=

a

8

and during deacceleration 0=8−a(4−t

1

) or t

1

=

a

8

∴8=a(4−

a

8

)

8=4a−8 or a=4 and t

1

=8/4=2sec

Now,Distance covered during acceleration s

1

=0×2+

2

1

×4(2)

2

or s

1

=8m

Distance covered during deacceleration s

2

=8×2−

2

1

×4×(2)

2

or s

2

=8m

Total distance =s

1

+s

2

=16m

Answer 2

Answer:

Benjamin knxn -_?:"-+:_-;-/_?+_+

Explanation:

sorry