1.

A car accelerates from rest at constant rest a alpha t for same time after which it deaccelerates at constant rate beta to come to rest. If total time taken is t then distance covered is given by :

Answer»

`X=t^(2)((alpha BETA)/(alpha+beta))`
`x=t^(2)((alpha beta)/(alpha-beta))`
`x=(t^(2))/(2)((alpha beta)/(alpha-beta))`
`x=(t^(2))/(2)((alpha beta)/(alpha+beta))`

Solution :x=distance covered during accelerated MOTION + disatnce covered during retarded motion
Let V be themaximum velocity REACHED.
`:. X=(v^(2))/(2alpha)+(v^(2))/(2beta)=v^(2)((2beta+2alpha)/(2alpha.2beta))`
Since `v=((alphabeta)/(alpha+beta))t:. x=((alphabeta)/(alpha+beta))^(2).t^(2).((alpha+beta))/(2alphabeta)`
`x=(1)/(2)((alphabeta)/(alpha+beta)).t^(2)`


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