That would be the aircraft with the highest thrust-to-weight ratio. This site lists the fighters of the world with a bit of a puzzling caveat for TO weight:
TWR or T/W ratio = (Max Thrust of Engine / (Empty Weight + (3.505 Tonnes of Fuel & Weapons, or only Internal Fuel)))
- 1.30 - Su-35BM
- 1.29 - F-15K
- 1.26 - Su-27S
- 1.25 - Eurofighter
- 1.24 - Mig-35
- 1.23 - Su-27SK & J-11A
- 1.19 - Rafale C
- 1.19 - Mig-29M/M2
- 1.19 - F-15C
- 1.18 - F-22 (T/W = 1.37 with Round nozzles)
Let's take the F-22 with the round nozzles, whatever those may be: thrust-to-weight of 1.37. Weight W = m * g, so
$$ a = \frac{T}{W} \cdot g = 1.37 * 9.81 = 13.44 ~\text{m/s}^2$$
60 mph = 26.82 m/s, and with V = a * t we get for t = V/a = 2.0 sec. The F-22 with the round nozzles reaches 60 mph in 2.0 seconds.
Or with TWR the thrust-to-weight ratio, and all SI units: $$ t = \frac{V}{g \cdot TWR}$$
Edit
The above is of course for frictionless circumstances, as @Manu H pointed out in a comment. Also, @Zeus pointed out that the "F-22 with the round nozzles" never became anything more than a gleam in some engineer's eye. So what do we get if we take the aircraft listed as highest T/W - the Su-35 - and make an attempt at a ROM for friction effects?
Let's take an SU-35 at TWR of 1.3. Resisting friction is caused by:
- Rolling friction of the tyres. This Wikipedia site lists rolling resistance coefficients, let's take 0.006 (about the minimal car tyre value), so tyre drag = 0.006 the weight of the aircraft.
- Aerodynamic friction. Let's assume that this is caused by lift-less drag $C_{D_0}$ only. This Wiki site gives subsonic drag coefficient of 0.021 for a Phantom, let's take that. This Wiki site gives further data on the Su35, such as max thrust and wing area. In order to match the TWR of 1.3 with the stated afterburner thrust of 284 kN, we'll take a mass of 22,269 kg.
The aerodynamic drag is a quadratic function of the velocity. At 60 mph = 26.82 m/s, the aerodynamic drag = $ C_{D_0} \cdot \frac{1}{2}\rho \cdot V^2 \cdot S$ = 0.021 * 0.5 * 1.225 * 62 * 26.82$^2$ = 574 N = 0.2% of thrust. That is at the end speed, the average weighted value is a third of that = 0.07% of thrust
Tyre drag = 0.006 * m * g = 1,311 N = 0.5% of thrust
So if we account for drag, we need to use about 99.4% of thrust. Time to reach 60 mph now becomes
$$ t = \frac{V}{g \cdot 0.994 \cdot TWR} = \frac {26.82}{9.81 \cdot 0.994 \cdot 1.3} = 2.1 ~\text{sec}$$
What a way to spend a Sunday...