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Let acceleration be a m/s^2 and initial velocity be u m/s. Now s(distance) in first part is 5 m and final velocity is 6m/s therefore, using- v^2-u^2=2as we get, 6^2-u^2=2*a*5 => 36-u^2=10a --------- eq.1 now in second part of journey...6m/s becomes initial velocity, 8m/s becomes final velocity and 7m is s again applying the same equation...we get, 8^2 - 6^2=2*a*7 => 64-36=14a =>28=14a =>a=2m/s^2 NOW SUBSTITUTING VALUE OF a IN EQ. 1, WE GET- 36-u^2=10*2 =>36-u^2=20 =>36-20=u^2 =>16=u^2 =>u=4m/s Therefore, we get- acceleration = 2m/s^2 and initial velocity= 4m/s