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“? (27m, 9 m/s)

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Answer 1

pls mark as brainlist answer

Answer 2

Answer:

Explanation:

v5=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