Correct answer is (20)

Distance travelled = Area under speed – time graph

= $\frac{1}{2}\times 8\times 5$ = 20 m

Correct Option is (D)

Given initial velocity u = 0 and acceleration is constant
At time t
v = 0 + at
$\Rightarrow$ v = at

Also $x=0(t)+\frac{1}{2}a{t}^2$
$\Rightarrow$ x = $\frac{1}{2}a{t}^2$

Graph (A), (B) and (D) are correct

Correct Option is (B)

S = Area under graph

$\frac{1}{2}$ $\times$ 2$\times$2 + 2 $\times$ 2 + 3 $\times$ 1 = 9 m

Correct Option is (C)

In option (A) you can see velocity versus time graph is a straight line with negative slope, so the acceleration is negative and constant. This motion is look like this.

Initially velocity is maximum and as acceleration is negative so after travelling some distance velocity will will become zero and then velocity will increase in the negative direction. At option (B) you can see same situation happens so (A) and (B) represent same situation.

From diagram, Initially at t = 0 position h = 0 and after some time h become maximum at the end h again become zero. Option (D) represent this situation.

So, A, B and D are correct.

The distance - time graph for this case will be -

But the given distance-time graph in the question does not match with this so option (C) will be wrong answer.

Correct Option is (A)

Till 15 sec car has accelerative motion and scooter has constant velocity in entire motion.

Total Distance travelled by the car in 15 sec, = $\frac{1}{2}$ $\times$ $\frac{45}{15}$ $\times$ (15)² = $\frac{675}{2}$ m

Distance travelled by scooter in 15 sec.

= V $\times$ t = 30 $\times$ 15 = 450 m

$\therefore$ Difference between distance travelled by the car and scooter in 15 sec,

= 450 $-$ $\frac{675}{2}$ = 112.5 m

Now, assume car catches the scooter in time t,

$\therefore$ Car travelled in t sec = scooter travelled in t sec.

$\Rightarrow$ $$ $\frac{675}{2}$ + 45(t$-$ 15) = 30 $\times$ t

$\Rightarrow$ $$ 337.5 + 45t $-$ 675 = 30t

$\Rightarrow$ $$ 15 t = 337.5

$\Rightarrow$ $$ t = 22.5 sec.