A car accelerates from rest at a constant rate α for some time . . .

Question : A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t, then calculate 

(a) the maximum velocity acquired by the car, and 
(b) Total distance travelled by the car.

Doubt by Ishmeet 

Solution : 
Case I :
let the body accelerates up to time time t1
Initial velocity (u) = 0 m/s
acceleration (a) = α
We know, 
v=u+at
v=0+αt1
v=αt1 — (1)

Case II :
let the body decelerated up to time t2
here initial velocity (u) = αt1
accelerartion (a) = -β
final velocity (v) = 0
Again using 
v=u+at
0=αt1+(-β)t2
-αt1=-βt2
αt1=βt2 — (2) 

Also 
t1+t2=t
t2=t-t1 — (3)
putting in eq (2)
αt1=β(t-t1)
αt1=βt-βt1
αt1+βt1=βt
t1(α+β)=βt
t1=βt/(α+β) — (4) 

putting in equation (3)
t2=t-[βt/(α+β)]
t2=[t(α+β)-βt]/(α+β)
t2=[αt+βt-βt]/(α+β)
t2=αt/(α+β) — (5) 

(i) From Equation (1) 
v=αt1
substituting the value of t1 from equation (4)
v=αβt/(α+β)
This the maximum velocity aquired by  the car. 

(b) Let distance covered by the car during acceleration = s1
Using 
s=ut+½at²
s1=0(t1)+½αt1²
s1=½α[βt/(α+β)]² [using eq (4)]
s1=αβ²t²/[2(α+β)²]

Let distance covered by the car during deceleration = s2
Again
s=ut+½at²
s=ut2+½at2²
s2=[αβt/(α+β)][αt/(α+β)]+½(-β)[αt/(α+β)]²
s2=α²βt²/(α+β)²-α²βt²/2(α+β)²
s2=α²βt²/2(α+β)²

Total distance covered = s1+s2
s=αβ²t²/2(α+β)² + α²βt²/2(α+β)²
s=[αβt²/2(α+β)²](β+α)
s=[αβt²/2(α+β)²](α+β)
s=αβt²/2(α+β)