#### Answer

The minimum acceleration is $4.4~m/s^2$

#### Work Step by Step

At a speed of 30 m/s, the train will reach the crossing in 2 seconds. In the first 0.50 seconds, the car moves 10 meters. Therefore, the car needs to go at least 35 meters in 1.5 seconds. We can find the minimum acceleration as;
$y = v_0t+\frac{1}{2}at^2$
$a = \frac{2y - 2v_0t}{t^2}$
$a = \frac{(2)(35~m) - (2)(20~m/s)(1.5~s)}{(1.5~s)^2}$
$a = 4.4~m/s^2$
The minimum acceleration is $4.4~m/s^2$.