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A body leaving a certain point O moves with an acceleration which is constant. At the end of the first second after it left O, its velocity is 1.5m/sec. At the end of the sixth second after it left O the body stops momentarily and begins to move backwards. The distance traveled by the body before it stops momentarily is

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Solution

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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+12at2

s1=9×6+12×(−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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