Question

A particle accelerates from rest at a constant rate for some time and attains a velocity of 8m/s. Afterwards it deaccelerates with the constant rate and comes to rest .If the total time taken is 4sec, what is the distance travelled?​

Answer 1

Explanation:

ANSWER

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:

Merry Christmas !!!

Explanation:

Good night :)

Have a great day ahead....