Question

A body starts from rest with uniform
acceleration. The velocity of the body
after t seconde is v. What is the displacement
of the body in the last three seconds?​

Answer 1

Answer:

The distance traveled by the particle in the last 2 seconds = distance traveled by particle in nth second- distance traveled by particle in (n-2)th second

So,

s=

2

1

an

2

−

2

1

a(n−2)

2

s=

2

1

a[n

2

−n

2

+4n−4]

s=

2

a

[4n−4]

s=(2n−2)a=2a(n−1).........(1)

Now,

Velocity, v=acceleration×time

v=a×n

a=

n

v

...........................(2)

Substitute value of 'a' from (2) to (1) we get,

s=

n

2v(n−1)

Answer 2

Answer:

Displacement s is 3v.

Explanation:

v=ds/dt

ds=vdt

integrating on both sides,

int ds= int (vdt)

int lim( 0-s) ds= v [int lim(0-3) dt]

s-0=v (3-0

s=3v

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