JEE Main 17 March 2021 Shift 1 Solved Paper

Consider 1st case,
A car starts from rest at a constant acceleration α. It means
Acceleration, a1=α
initial velocity, u=0
v=u+a1t⇒v0=0+αt1
⇒‌‌v0=αt1...(i)
Consider 2nd case,
The same car decelerates after some time at a constant acceleration β and comes to rest. It means
Acceleration, a2=β
Final velocity =0
⇒‌‌v0=βt2...(ii)
According to question,
t1+t2=t
⇒‌‌v0(‌

1

α

+‌

1

β

)=t  [from Eqs. (i) & (ii)]
⇒‌‌v0=‌

αβt

α+β

...(iii)
If we draw the velocity-time graph for the given situation, it will be as follows∴ Distance travelled = Area of v−t graph
=‌

1

2

×t×v0=‌

1

2

×t×‌

αβt

(α+β)

 [from Eq. (iii)]
=‌

1

2

‌

αβt2

(α+β)

=‌

αβt2

2(α+β)