A car accelerates from rest at a constant rate ` alpha ` for some time after which it decelerates at a constant rate ` beta` to come to rest. If the total time lapse is ` t` seconds , evauate.
(i) maximum velocity reached , and
(ii) the total distance travelled .

Explanation

`t_(1)+t_(2)=t_(0),t_(2)=t_(0)-t_(1)`
`v_(m)`: Maximum velocity
v-t graph:
`A` to `C`: `v=u+a t=0+alphat=alphat`
`t=0, v=0`
`t=t_(1), v=v_(m)`
Shape: `y=alphax` (straight line)

`C` to `B`: `v=u+at=v_(m)-betat`
`t=t_(1), v=v_(m)`
`t=t_(0), v=0`
Shape: `y=v_(m)-betax` (straight line)
`y=c+mx`
`c=v_(m)+ve`
`m=-beta, -ve`

A to B:

(a) Slope of the v-t graph gives acceleration
A to C: `v_(m)/t_(1)=alphaimpliest_(1)=v_(m)/alpha ...(i)`
C to B: `v_(m)/(t_(0)-t_(1))=betaimpliest_(0)-t_(1)=v_(m)/beta ...(ii)`
Adding (i) and (ii), we get
`t_(0)=v_(m)/alpha+v_(m)/betaimpliesv_(m)=(alphabetat_(0))/(alpha+beta)`
(Thought this part of the problem can be solved easily by analytical method, but here we want to show the procedure by graphical method.)
(b) Area of the v-t graph gives the displacement//distance.
`t=0 to t=t_(0)`, Area`=1/2 v_(m)t_(0)`
`=1/2(alphabetat_(0)^(2))/((alpha+beta))`=Distance