Question

A body leaving a certain point " O" moves with a constant
acceleration. At the end of the 5 th second its velocity is 1.5
m/s. At the end of the sixth second the body stops and then
begins to move backwards. Find the distance traversed by the
body before it stops. Determine the velocity with which the
body returns to point " O " ?

Answer 1

Answer:

v=u+at

1.5=u+5a...........(i)

0=u+6a.............(ii)

From (i) and (ii)

0=1.5+a(6−5)

a=−1.5m/s

2

Om substituting a=−1.5m/s

2

in equation (ii)

0=u+6(−1.5)

0=u−9

u=9m/s

The distance traveled by the body in its forward journey is s

1

can be calculated as follows,

s

1

=ut+

2

1

at

2

s

1

=9×6+

2

1

×(−1.5)×6

2

s

1

=54+(−1.5×18)

s

1

=27m

The distance traveled by the body in its forward journey (s

1

)= the distance traveled by the body in its backward journey(s

2

)=27m

So, the total distance traveled by the body in forward journey and its backward journey=s

1

+s

2

=27+27=54m

Answer 2

Hello mate..

Here Is Your Answer:

GIVEN:

• Velocity in 5th sec = 1.5 m/s
• The body comes to rest in 6th sec.

Therefore,

Final velocity in 6th sec, v = 0.

Acceleration in 6th sec is v = u + at

=> 0 = 1.5 + a × 1

=> a = -1.5 m/s²

[The velocity in 5th second becomes the initial velocity for 6th second]

After 6 seconds, the body comes to rest.

Therefore,

v = 0,

a = -1.5 m/s²

u = ?

t = 6 sec.

$\bold{v = u + at}$

=> 0 = u - 1.5 × 6

=> u = 9 m/s.

Therefore,

Distance traversed by the body in 6 seconds .i.e., before it stops.

$\bold{s = ut + \frac{1}{2} {at}^{2} }$

=> 9 × 6 + ½ × -1.5 × 6²

=> 54 - 27 = 27m.

For backward journey,

u = 0 m/s

t = 6 seconds

a = -1.5 m/s²

$\bold{v = u + at}$

=> v = 0 - 1.5 × 6

=> v = -9

Therefore, Velocity for backward journey is -9 m/sec.

⠀

HOPE THIS HELPS YOU..