Special Relativity, change in distance and time at slow speeds

[kambrian]

I was able to do pretty much every other problem dealing with this topic except this one which is bothering me. Here is the question:

The distance from New York to Los Angeles is about 5000 km and should take about 50 h in a car driving at 100 km/h. (a) How much shorter than 5000 km is the distance according to the car travelers? (b) How much less than 50 h do they age during the trip?

Can someone please give me a hint as to which equation I am looking for to solve this problem? Basically all equations with sqrt(1-(u^2/c^2)) are coming out completely wrong for me.

I am able to get to sqrt(1-(27.78^2/3x10^8)) but don't really know how to deal with subtracting such a small number. Obviously my calculator is throwing 1 at me which screws up the equation.

Any help is appreciated.

 

[Doc Al]

Use the binomial expansion of (1 + x)^n (where x << 1) to estimate the change in distance: (1 + x)^n = 1 + nx + ...

 

[kambrian]

I am still at a loss here. I found the expansion to be 1+x/2-(x^2)/8 which really does not help me much. With x = 8.57*10^-15, I am still getting what essentially is 1.

 

[Janus]

Do you have a scientific calculator function on your computer? (With Windows, it should be under "accessories".) I know that the one on my computer is capable of handling that many decimal places.

 

[Doc Al]

I am still at a loss here. I found the expansion to be 1+x/2-(x^2)/8 which really does not help me much. With x = 8.57*10^-15, I am still getting what essentially is 1.

Realize that you are being asked for the change in length, not the length.

Try this:
$$L = L_0 \sqrt{1 - v^2/c^2}= L_0 (1 - v^2/c^2)^{1/2}$$

Thus, using a binomial expansion (for v^2/c^2 << 1) :
$$L \approx L_0 (1 - (1/2)v^2/c^2)$$

So:
$$\Delta L \approx -(1/2) L_0 v^2/c^2$$