1. Jan 25, 2004

KingNothing

Can anyone confirm/deny that if an object is dropped (init. velocity=0) that it will take about 2.26 seconds to reach 90 km/h?

$$S=\frac{D}{T}$$
$$A=\frac{V_f-V_i}{T_f-T_i}$$
$$D=\frac{1}{2}gT^2$$
$$V_f^2=V_i^2+2a(X_f-X_i)$$
$$\overline{V}=\frac{V_f+V_i}{2}$$

Last edited: Jan 25, 2004
2. Jan 25, 2004

Jimmy

Assuming no air resistance and an acceleration of 9.81 m/s^2, I get 2.55 seconds.

Or if the object falls for 2.26 seconds, it will have a velocity of 80 Km/hr.

Last edited: Jan 25, 2004
3. Jan 25, 2004

KingNothing

Can you go through it for me? It's really confusing me as to why I keep getting what I do.

4. Jan 25, 2004

Jimmy

Time = Velocity/acceleration

The units I used were Seconds, m/s, m/s^2

I converted your value of 90 km/hr to m/s. Divide by 3.6 to convert km/hr to m/s. 90 km/hr = 25 m/s

I simply plugged in the values into the above formula.

T = 25 /9.81 = 2.55 seconds.

Your answer wans't really that far off. What values did you use?

5. Jan 25, 2004

KingNothing

Just to start off, we have to assume gravity to be 9.8 in our physics class, but I don't think that can attribute to the difference. Anyhoo, what is wrong with using:

$$V_f = V_i + \frac{1}{2} a t^2$$
If $$V_f$$=25 and $$\frac{1}{2}a$$=4.9 and $$V_i$$=0, then what should I come out with, or is this the wrong equation?

Last edited: Jan 25, 2004
6. Jan 25, 2004

Jimmy

It's the wrong equation. $1/2 a t^2$ is an expression of distance traveled under a constant acceleration for a given length of time.

Since $V_i=0$, your equation simplifies to $V_f = 1/2 a t^2$ which is not correct. $D = 1/2 a t^2$ when intial distance and velocity equal 0. Otherwise,

$$D = D_i+V_i t + \frac{a t^2}{2}$$

the units on one side of the equation have to match the units on the other side. That's a good way to check for errors in an equation.

$$V_f = V_i + a t$$

Last edited: Jan 25, 2004