# Constant Acceleration Car PRoblem

1. Oct 14, 2007

### Thiendrah

1. The problem statement, all variables and given/known data
You're driving at 70 km/h when you accelerate with constant acceleration to pass another car. Six seconds later, you're doing 80 km/h.

How far did you go in this time?

2. Relevant equations
V= V(0)+at
V^2- V(0)^2 = 2a ( X-X(0) )

3. The attempt at a solution

I keep finding some large numbers like 125000 m. I think i'm doing something wrong with converting times. So can someone tell me how to do this problem?

2. Oct 15, 2007

### Staff: Mentor

3. Oct 15, 2007

### Thiendrah

you kinda confused me.

That's what i did. I found the acceleration from the first equation. V=V(0)+at. I mean we know V=80km/h, V(0)= 70km/h, and t=6. Only thing left is a, right?. If i plug the acceleration i found with this in second equation. The answer isn't correct.

4. Oct 15, 2007

### Staff: Mentor

make sure one does not mix hours and seconds in the acceleration.

The v is given in km/h and the time in seconds.

One could certainly write the acceleration in km/h/s, but then when using v2 one would have units of (km/h)2.

I recommend converting km/h to m/s first.

It is better to show the work and make sure units are consistent.

5. Oct 15, 2007

### Thiendrah

Ok, here is a weird thing.

The site you gave me. I found the average velocity. Then that times with 6s gave me the displacement. which is the answer.

But the thing is i can't find the answer with those equations. It's basically the same thing. Still can't find it.

6. Oct 15, 2007

### CompuChip

What you did is right, but you should watch the units. I repeated your calculation but put in the units...
For example (6 seconds = 60/10 seconds = 1/10 hour):
$$a = \frac{V - V_0}{t} = \frac{10\text{ km/h}}{1/10\text{ h}} = 100 \text{ km/h}^2.$$

Then from the second equation,
$$d \equiv (X - X_0) = \frac{ V^2 - V_0^2 }{ 2 a } = \frac{ \left( 80^2 - 70^2 \right) (\text{ km/h})^2 }{ 2 \times 100 \text{ km/h}^2 } = \frac{1500}{200} \frac{\text{km}^2 /\text{h}^2 }{ \text{km} / \text{h}^2 } = \frac{15}{2} \text{ km},$$
which IMO is still a lot (are the formulas right, did I make any calculation errors?) but better than 125 km

See how it works? Alternatively, you could have converted everything to m and s.

7. Oct 15, 2007

### Thiendrah

Only thing i did wrong was converting units. Everytime i did this problem, I'd have numerator and denominator in different units for the first equation.

Thanks both.

8. Oct 15, 2007

### Staff: Mentor

6 sec = 0.1 minute = 0.1/60 h or 6/3600 h = 0.00167 h ( 1 sec = 1/60 min = 1/3600 hr)

The delta-V = 80 km/h - 70 km/h = 10 km/h = 10000m/3600s = 2.78 m/s

dV/dt = 2.78 m/s / 6 s = 0.463 m/s2

or

10 km/h/ (0.00167 h) = 5988 km/h2

So one has to be careful with units.

9. Oct 15, 2007

### CompuChip

Hehe, of course!
I thought there was a mistake somewhere.
Seven kilometers in 6 seconds seemed like a bit much
Thanks.