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 :

Options

A. `x=t^(2)((alpha beta)/(alpha+beta))`
B. `x=t^(2)((alpha beta)/(alpha-beta))`
C. `x=(t^(2))/(2)((alpha beta)/(alpha-beta))`
D. `x=(t^(2))/(2)((alpha beta)/(alpha+beta))`

Answer

d

Explanation

x=distance covered during accelerated motion + disatnce covered during retarded motion
Let v be the maximum 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)`