Question is: a car travelling at 140 km/h decelerates for 20 secs. During this time it travels 500 m. What is its deceleration? What is its speed after deceleration?

Hello!
As I understand, the deceleration is uniform (the same all the time). Denote it as agt0 and denote the initial speed as V_0.  In m/s V_0 = 140/3.6.
Then the speed is V(t) = V_0 - a*t  (note the minus sign), and the displacement is D(t) = V_0 t - (a t^2)/2.
It is given that D(20) = 500, therefore V_0 * 20 - (a * 20^2)/2 = 500. From this we find  a = 2*(140/3.6*20-500) / 20^2 approx1.39 (m/s^2).
Then we can find the speed after deceleration: it is
V(20) = V_0 - a*20 approx 140/3.6 - 20*1.39 approx 11.1 (m/s).
The value is positive which means the direction of movement remains the same.
The answers: the deceleration is about 1.39 m/s^2, and the final speed is about 11.1 m/s.