A body covers 20 m distance …
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Sahdev Sharma 6 years, 9 months ago
Let initial velocity = u
Acceleration = a
Using equation
{tex}s = ut+{1\over 2}at^2{/tex}
For first 2 sec
{tex}20 = 2u+{1\over 2}4a{/tex}
{tex}20 = 2u+2a{/tex}
{tex}10=u+a \ ....... (1){/tex}
Now distance covered in 2 to 5 sec is given 40m
{tex}40 = 5u+{1\over 2}25a-20{/tex}
=> 120 = 10u +25 a
=> 24 = 2u+5a ....(2)
Solving (1) and (2), we get
{tex}a= {4\over 3}\ m/s^2{/tex}
And u {tex}={26\over 3}\ m/s{/tex}
2Thank You