Question

hey guys tell me how it came

Answer 1

My dear your easiest solution is here!!!!!!!!!!!!! Hey I think this solution of your answer might make u understand properly!! :-) :-) :-) :-)
First, Try to give full focus and write it down in your copy so that every symbol used on this u can catch easily!!!!!! Regarding any minor mistaken comment in box!!!!!
Solution)----On simplest step wise------
(@) --The position at time t is..
or position1 i.e. s=ut+1/2×at^2(using eq.1)
and The position at time (t-1s) is
or position2 i.e. s'=u(t-1s) +1/2×a(t-1s) ^2 {step1}
On expending (t-1s) ^2as (-b) ^2.We get,,
=u(t-1s) +a/2{(t)^2-2t(1s) +(1s)^2} {step2}
=u(t-1s) +a/2×t^2-at(1s) +a/2(1s)^2 {step3}
=ut-u(1s) +a/2×t^2-at(1s)+a/2(1s)^2{step4 Thus, we have the displacement in the last 1s is $t=(ut+1/2at^2)-(ut-u(1s)+1/2at^2-at(1s)+a/2(1s)^2
=ut+1/2at^2-ut+u(1s) -1/2at^2+at(1s) -a/2(1s)^2
(After solving it -----the new eq. Forms as --
=u(1s) +at(1s)-a/2(1s)^2
=u(1s) +{at(1s) -a/2(1s)^2} [On taking common] we get, =u(1s)+a/2(2t-1s)(1s) -------(1)ans. Of (a)
Now.. Put this value in question (b) as u know already!
Hope Certainly It helps you my dear !!!!!! :-) :-) :-)