You have to split the time domains into the gears needed to reach 60mph. For each gear, there have to be assumptions on the power delivery of the car.
Typically 1st gear is traction limited, so you can assume constant acceleration up to the speed where peak power occurs. The relationship between power speed and acceleration is $P(v) = m \,v\, a(v)$. So run constant acceleration equal to traction of $a_1 = \epsilon \, g$ to speed $v_1 = \frac{P_{max}}{\epsilon g m}$ where $\epsilon$ is traction coefficient (0.4 for FWD, 0.6 for RWD and 0.9 for AWD) representing the peak 'g loading in 1st gear. So in the end of 1st gear the car has parameters
$$ \begin{align}
t_1 & = \int_0^{v_1} \frac{1}{a(v)}\,{\rm d} v= \frac{ \frac{P_{max}}{m} }{\epsilon^2 g^2} \\
x_1 & = \int_0^{v_1} \frac{v}{a(v)}\,{\rm d} v=\frac{ \left(\frac{P_{max}}{m}\right)^2 } {2 \epsilon^3 g^3} \\
v_1 & = \frac{ \frac{P_{max}}{m} } {\epsilon g} \\
a_1 & = \epsilon g
\end{align} $$
This is the easy part. Now for the sprint to 60mph. Also use the parameter $w = \frac{P_{max}}{m}$ for the power to weight ratio. From this point on, the acceleration is a function of speed and it can be either:
$$\begin{aligned}
\mbox{no air resistance} & & a(v) & = \frac{w}{v} \\
\mbox{with air resistance} & & a(v) & = \frac{w}{v} - \beta v^2 \\
\end{aligned}$$
The math is simper without air resistance to calculate the parameters for $v_{60} = 60 {\rm mph}$
$$ \begin{align}
t_2 & = t_1 + \int_{v_1}^{v_2} \frac{1}{a(v)}\,{\rm d} v= \frac{w}{2 \epsilon^2 g^2} + \frac{v_{60}^2}{2 w} \\
x_2 & = x_1 +\int_{v_1}^{v_2} \frac{v}{a(v)}\,{\rm d} v= \frac{w^2}{6 \epsilon^3 g^3} + \frac{v_{60}^3}{3 w}\\
v_2 & = v_{60} \\
a_2 & = \frac{w}{v_2} \\
\end{align} $$
So the time to 60 can be estimated as
$$\boxed{ t_{60} = \frac{P_{max}}{2 m \epsilon^2 g^2} + \frac{m v_{60}^2}{2 P_{max}} } $$
Example
A $m=1200\,{\rm kg}$ car with peak power $P_{max} = 160\,{\rm hp} = 119,000\,{\rm W}$ goes to $v_{60} = 26.9\,{\rm m/s}$. Traction is $\epsilon=0.4$ and $g=9.81\,{\rm m/s^2}$
$$ t_{60} = \frac{ \frac{119,000}{1200} }{2 \times 0.4^2 \times9.81^2} + \frac{26.9^2}{2 \frac{119,000}{1200}} = 3.23 + 3.63 = 6.86 \, {\rm sec} $$
The actual numbers are going to be slower, since less than peak power is delivered most of the time, and there is air resistance and rolling resistance, plus time to shift gears and road grade effects and ...
Edit 1
For an electric car, given a linear torque curve $T(rpm) = T_{max} \left(1 - \frac{rpm}{rpm_{max}} \right)$ you can create a function of power to weight as function of speed $$w(v) = \frac{P(v)}{m} = \frac{v T(v)}{m} = v( C_0 - C_1 v)$$ given the gearing $rpm(v) = \gamma v$
Now use the acceleration $$ a(v) = \frac{w}{v} - C_2 v^2 = \frac{v (C_0 - C_1 v)}{v} - C_2 v^2 = C_0 - C_1 v - C_2 v^2 $$
With direct integration you have
$$ t_1 = \int_0^{v_1} \frac{1}{a}\,{\rm d}v = \int_0^{v_1} \frac{1}{C_0-C_1 v-C_2 v^2}\,{\rm d}v = \ldots$$
With the parameter of top speed $a(v_f) = 0 \} v_f = \dfrac{\sqrt{C_1^2+4 C_0 C_2}-C_1}{2 C_2} $ and the dimensionless parameter $\zeta = 2-\frac{C_1 v_f}{C_0}$ the time to speed is
$$ \boxed{ t(v) = \frac{v_f}{C_0 \zeta} \ln \left(1+\zeta \frac{v}{v_f-v}\right) }$$
NOTE: There is constraint that says at terminal speed the motor must make positive torque, or the drag limited top speed must be less than the rpm limited top speed.
Edit 2
For constant torque $a = a_0 \left( 1 - \left( \frac{v}{v_f} \right)^2 \right)$ and so
$$ \begin{align} t & = \int\limits _{0}^{v}\frac{1}{a_{0}\left(1-\left(\frac{v}{v_{f}}\right)^{2}\right)}\,{\rm d}v\\
& = \frac{v_f}{a_0} {\rm tanh}^{-1} \left( \frac{v}{v_f} \right) \\
& = \frac{v_f}{2 a_0} \ln \left( \frac{v_f+v}{v_f-v} \right)
\end{align}$$
where $a_0$ is the initial acceleration (at zero speed) and $v_f$ the drag limited top speed.
