A Car Starts from rest and travelled 100m in the first 10s. Assuming uniform acceleration, Calculate the distance it will travel in the next 10s.

The car starts from rest, so its initial velocity $u = 0 \, \text{m/s}$.
It travels a distance $s_1 = 100 \, \text{m}$ in time $t_1 = 10 \, \text{s}$.
Using the kinematic equation $s = ut + \frac{1}{2}at^2$:

$$100 = (0)(10) + \frac{1}{2}a(10)^2$$ $$100 = 0 + \frac{1}{2}a(100)$$ $$100 = 50a$$ $$a = \frac{100}{50}$$ $$a = 2 \, \text{m/s}^2$$

The car travels for the first $10 \, \text{s}$ and then for the next $10 \, \text{s}$, making a total time $t_2 = 10 \, \text{s} + 10 \, \text{s} = 20 \, \text{s}$.
Using the same initial velocity $u = 0 \, \text{m/s}$ and the calculated acceleration $a = 2 \, \text{m/s}^2$:

$$s_2 = ut_2 + \frac{1}{2}at_2^2$$ $$s_2 = (0)(20) + \frac{1}{2}(2)(20)^2$$ $$s_2 = 0 + 1(400)$$ $$s_2 = 400 \, \text{m}$$

This is the difference between the total distance traveled in $20 \, \text{s}$ and the distance traveled in the first $10 \, \text{s}$.
Distance in next $10 \, \text{s} = s_2 - s_1$
Distance in next $10 \, \text{s} = 400 \, \text{m} - 100 \, \text{m}$
Distance in next $10 \, \text{s} = 300 \, \text{m}$

The distance the car will travel in the next $10 \, \text{s}$ is $\boxed{300 \, \text{m}}$.