Question

A body leaving a certain point O moves with an acceleration ,which is constant in magnitude and direction.At the end of fifth second ,its velocity is 1.5m/s.At the end of sixth body stops and then begins to move backwards.a)Find the distance travelled by the body before it stops .b)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/s2

Om substituting a=−1.5m/s2 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 s1 can be calculated as follows,

s1=ut+21at2

s1=9×6+21×(−1.5)×62

s1=54+(−1.5×18)

s1=27m

The distance traveled by  the body in its forward journey (s1)= the distance traveled by the body in its backward journey(s2)=27m

So, the total distance traveled by the body in forward journey and its backward journey=s1+s2=27+27=54m

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Answer 2

$\huge\underline\bold\red{AnswEr}$

v = u + at

1.5 m/s = u + 5a  ------------------ (1)

0 m/s = 1.5 m/s + a 

⇒ a = − 1.5 m/s2

Substituting the value of a in Eq. (1),

1.5 m/s = u + 5 x ( − 1.5 m/s2 )

⇒ u = 9 m/s

Distance travelled in 6 s (i.e. before it stops) S = ut + ½ at2

                                                                                                  = 9 x 6 + ½ ( − 1.5 m/s2 ) x 6 x 6

                                                                            = 54 - 27 = 27 m

On the return journey we know that initial velocity u = 0, distance covered S = 27 m. 

To determine the velocity with which the body returns to point o, we need the time taken or the acceleration in the backward direction. The we can substitute in v = u + at and get the answer.