Using relativistic velocity addition with c = 82 m/s, what is the speed of an oncoming truck relative to you when (a) you drive at 25 m/s and the truck drives at 35 m/s, and (b) you drive at 5.0 m/s and the truck drives at 55 m/s (all speeds relative to the ground)?

What we are trying to find

The truck and you are moving toward each other, so the “truck’s speed relative to you” means the truck’s velocity measured in your rest frame. Because $c$ is only $82\,\text{m/s}$ here, we cannot just add speeds classically.

Choose a sign convention (this is the key step)

Let the positive $x$-direction be the direction you drive.

• Your velocity relative to the ground: $v = +v_{\text{you}}$
• Truck velocity relative to the ground (oncoming): $u = -v_{\text{truck}}$

Apply the relativistic velocity transformation

To transform an object’s ground-frame velocity $u$ into your frame moving at $v$: $$ u' = \frac{u - v}{1 - \frac{uv}{c^2}} $$ The speed you observe is $|u'|$.

(a) $v_{\text{you}}=25\,\text{m/s}$, $v_{\text{truck}}=35\,\text{m/s}$

Set $c=82\,\text{m/s}$, $v=+25$, and $u=-35$: $$ u' = \frac{-35 - 25}{1 - \frac{(-35)(25)}{82^2}} = \frac{-60}{1 + \frac{875}{6724}} = \frac{-60}{1.13012} \approx -53.1\,\text{m/s} $$ So the truck’s speed relative to you is: $$|u'| \approx 53.1\,\text{m/s}.$$

(b) $v_{\text{you}}=5.0\,\text{m/s}$, $v_{\text{truck}}=55\,\text{m/s}$

Now $v=+5.0$ and $u=-55$: $$ u' = \frac{-55 - 5.0}{1 - \frac{(-55)(5.0)}{82^2}} = \frac{-60}{1 + \frac{275}{6724}} = \frac{-60}{1.04089} \approx -57.6\,\text{m/s} $$ So the truck’s speed relative to you is: $$|u'| \approx 57.6\,\text{m/s}.$$

Quick reasonableness check

Both relative speeds are less than $c=82\,\text{m/s}$, as special relativity requires. Also, they are smaller than the classical sum $60\,\text{m/s}$ because relativistic addition reduces the combined speed when speeds are a noticeable fraction of $c$.