At the instant, the traffic light turns green, a car starts with a constant acceleration of $2.2\,m/{s^2}$. At the same instant a truck traveling with a constant speed of $9.5\,m/s$, overtakes and passes the car.
(a) How far beyond the starting point will the car overtake the truck?
(b) How fast will the car be traveling at the instant?

Hint: At the instant when the car overtakes the truck the distance traveled by car and the truck is the same and the time has taken is also the same. Using the second equation of motion we can find the displacement of the car by equating it with the displacement of trucks we can find the time taken. Using that we can find the distance traveled. And the velocity of the car can be found using the first equation of motion.

Complete step by step answer:
It is given that the acceleration of a car is
$a = 2 \cdot 2\,m/{s^2}$
The car starts from rest, so the initial velocity of the car is zero.
$ \Rightarrow u = 0$
At the same instant, the car starts and a truck overtakes the car.
The velocity of the truck is given as $u' = 9.5\,m/s$
(a). We need to find the distance at which the car will overtake the truck.
We know that from the second equation of motion.
$s = ut + \dfrac{1}{2}a{t^2}$
On substituting the given values, we get,
$\Rightarrow s = 0 + \dfrac{1}{2} \times 2.2 \times {t^2}$
$\Rightarrow s = 1.1\,{t^2}$
This is the distance covered by the car. Since this is also the distance covered by the truck at the same time $t$.
Since we know velocity is equal to distance divided by time taken,
$v = \dfrac{s}{t}$
We can write the distance s as ,
$\Rightarrow s = v \times t$
That is in case of train we can write,
$\Rightarrow s = u' \times t$
Substituting the value of velocity of train, we get
$\Rightarrow s = 9.5t$
Now let us equate equation (1) and (2)
$ \Rightarrow 1.1{t^2} = 9.5t$
$\Rightarrow t = \dfrac{{9.5}}{{1.1}} = 8.6\,s$
Substituting the value of time in equation 2 we get
$\Rightarrow s = 9.5 \times 8.6$
$\Rightarrow s = 81.7\,m$

Therefore, the distance covered is $81.7\,m$

(b). Now let us find the velocity of the car.
For this let us use the first equation of motion in the case of the car.
$v = u + at$
Since $u = 0$ we get
$\Rightarrow v = 2.2 \times 8.6$
$ \Rightarrow v = 18.92\,m/s$

Therefore the velocity of the car at the instant when it overtakes the train is $v = 18.92\,m/s$.

Note:
Remember that the car starts when the light turns green so the initial velocity is zero since it starts from rest. The truck meets the car at this starting point. The next time when the car meets the truck both the truck and the car have traveled the same distance at the same time.