Answer: a moving clock runs slow: \(\Delta t = \dfrac{\Delta t_0}{\sqrt{1 - v^{2}/c^{2}}}\). For a particle at 0.7c, rest mass : moving mass = \(\sqrt{1-0.49} \approx 0.714\), i.e. about 0.71 : 1 (or 1 : 1.40).
Concept
Time dilation means the time interval between two events measured by an observer in relative motion is longer than the interval measured by an observer at rest relative to the events (the proper time \(\Delta t_0\)). In other words, moving clocks appear to run slow.
Derivation
Let a clock be at rest at position \(x'\) in frame S′, which moves with velocity \(v\) along x relative to frame S. It ticks at times \(t_1'\) and \(t_2'\), so the proper time interval is \(\Delta t_0 = t_2' - t_1'\).
The inverse Lorentz transformation for time is \(t = \gamma\left(t' + \dfrac{vx'}{c^{2}}\right)\), with \(\gamma = \dfrac{1}{\sqrt{1 - v^{2}/c^{2}}}\). So
\[t_2 - t_1 = \gamma\left[(t_2' - t_1') + \frac{v}{c^{2}}(x' - x')\right]\]
Both ticks happen at the same \(x'\), so the second term is zero:
\[\Delta t = \gamma\,\Delta t_0 = \frac{\Delta t_0}{\sqrt{1 - v^{2}/c^{2}}}\]
Since \(\gamma > 1\), \(\Delta t > \Delta t_0\). Experimental proof: fast-moving muons from cosmic rays live long enough to reach the Earth's surface.
Numerical (\(v = 0.7c\))
Relativistic mass: \(m = \dfrac{m_0}{\sqrt{1 - v^{2}/c^{2}}}\), so
\[\frac{m_0}{m} = \sqrt{1 - (0.7)^{2}} = \sqrt{0.51} \approx 0.714\]
The moving mass is about 1.4 times the rest mass.
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