A car starts running at an initial speed of $60\text{km/h}$, with its speed increasing every hour by $10\text{km/h}$. How many hours will it take to cover a distance of $1210\text{km}$?

The distance covered in each consecutive hour forms an Arithmetic Progression (AP):$1^{\text{st}}$ hour distance = $60\text{km}$$2^{\text{nd}}$ hour distance = $70\text{km}$$3^{\text{rd}}$ hour distance = $80\text{km}$ Here, the first term $a=60$ and the common difference $d=10$.Let the total time taken be $n$ hours. The sum of the distances covered in $n$ hours is given by the sum formula of an AP ($S_n$):
$$S_n=\frac{n}{2}[2a+(n-1)d]$$
Given that $S_n=1210\text{km}$:
$$1210=\frac{n}{2}[2(60)+(n-1)10]$$
$$2420=n[120+10n-10]$$
$$2420=n[110+10n]$$
$$2420=110n+10n^2$$
Dividing the entire equation by $10$:
$$n^2+11n-242=0$$
Now, we solve this quadratic equation by splitting the middle term:
$$n^2+22n-11n-242=0$$
$$n(n+22)-11(n+22)=0$$
$$(n-11)(n+22)=0$$
Since time cannot be negative, $n=11$.Therefore, it will take $11$ hours to cover the distance.