Formula for trajectory also applicable to decelerating bodies?

[ionowattodo]

The formula that is applied to falling objects with disregard to any exterior forces except gravity is : d=0.5(g)(t^2)

Can it also be applied to decelerating bodies, such as cars, with disregard to friction?
Such as a car with an initial velocity slowing down to 0 m/s over a distance of 50 meters.

 

[diazona]

No, that formula only works for a body in free fall, starting from rest, in Earth's gravity (or equivalent).

For other accelerating or decelerating objects, there are various formulas you can use:
$$d = v_i t + \frac{1}{2}a t^2$$
if you know initial speed and acceleration
$$d = v_f t - \frac{1}{2}a t^2$$
if you know final speed and acceleration
$$d = \frac{(v_f + v_i)t}{2}$$
if you know initial and final speeds and time
$$d = \frac{v_f^2 - v_i^2}{2a}$$
if you know initial and final speed and acceleration

Of course, all these are only applicable when the acceleration is constant.

 

[ionowattodo]

Can't you substitute gravity for deceleration?
Basically if you turn a deceleration car to it going upwards, it's the same as a thrown ball except with different deceleration rates

 

[diazona]

Well, yes, it's the same, but the rate of acceleration/deceleration is not g for a car. The formulas I gave in my post work for any constant acceleration/deceleration. But the one you were originally asking about is what you get when the acceleration happens to be equal to 9.8 m/s^2 (and the initial speed is zero).