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).

 

[PJC01]

The formula for trajectory, d=0.5(g)(t^2), is specifically designed to calculate the distance traveled by a falling object under the influence of gravity. It takes into account the acceleration due to gravity (g) and the time (t) the object has been falling. Therefore, it may not be applicable to decelerating bodies such as cars, as they are not solely affected by gravity but also by other external forces such as friction and air resistance.

To accurately calculate the distance traveled by a decelerating body, we would need to consider these additional forces and incorporate them into the formula. A more appropriate formula for a decelerating body would be d=V0t + 0.5at^2, where V0 is the initial velocity, a is the deceleration rate, and t is the time.

In the example given, the distance traveled by the car would be calculated using this formula as d= (0 m/s)(t) + 0.5(-a)(t^2) = -0.5at^2. This formula takes into account the initial velocity of 0 m/s and the deceleration rate (a) over a distance of 50 meters.

In conclusion, while the formula for trajectory may be applicable to falling objects, it cannot be used to accurately calculate the distance traveled by decelerating bodies such as cars. A different formula that takes into account all the relevant forces would need to be used for such calculations.