What you've given are performance specifications, not really a full drive cycle, so you can do some analysis on requirements based on those performance specs but you can't, for example, use those performance specifications to specify the battery pack that should be used because the drive cycle isn't given. You'd need to specify how many times it should accelerate/decelerate, how long it operates at each speed, how many ramps it climbs and descends, how fast it operates on those ramps, etc.
For developing product specifications based on performance specs, you need to evaluate the product requirement for each use case, and then you take the product specification from the most demanding use case. From your specifications:
Vehicle top speed. This should be a negligible demand because it only requires overcoming the rolling friction in the wheels and other drivetrain losses.
Acceleration. Need to figure out the force required to get from 0 to 25 mph in the course of 60 feet (lowest acceleration). To do that:
Starting speed is zero, ending speed is 25 mph (36.67 feet per second). If there's constant acceleration then it must move at an average speed of 18.335 feet per second. To cover 60 feet, it must take (60/18.335)=3.27 seconds.
Considering the constant acceleration equation:
$$
x_f = x_0 + v_0t + \frac{1}{2}at^2 \\
$$
Assuming the vehicle starts stationary ($v_0=0$) at position $x_0=0$ then, for constant acceleration,
$$
x_f = \frac{1}{2}at^2 \\
\frac{2x_f}{t^2} = a \\
\frac{2(60 \text{ft})}{3.27^2 \text{s}^2} = a \\
\frac{120}{10.69} \text{ft/s}^2 = a \\
\boxed{a = 11.23 \text{ft/s}^2}
$$
Given Imperial units, gravity on Earth is $\approx 32.2 \text{ft/s}^2$, so you're looking at an acceleration of (11.23/32.2) = 0.35g.
- Climbing an incline. You say
the vehicle must be able to achieve top speed, which is a little unclear, because I can't tell if it should just maintain top speed on the incline, or if it should accelerate to top speed on the incline. I'll assume it's just maintaining the top speed up the incline. This goes back to point 1, where maintaining the speed should be trivial; the real work happens to prevent the "falling" force down the incline.
You already linked my other, related answer, so I'll just skip to the punchline there - the force you need to go up an incline is $mg\sin{\theta}$.
We can compare the acceleration case with the incline-climbing case if we regroup the result as $m\left(g\sin{\theta}\right)$. At 10 degrees (0.1745 radians), $\sin{\left(10^{\circ}\right)}g$ = 0.17g.
So, in looking at this, your easiest acceleration case requires 0.35g and maintaining speed on an incline requires 0.17g, so acceleration is the limiting case here - any motor capable of meeting your acceleration specification will be sufficient to also maintain speed up a 10 degree ramp.
Now the question is, how big of a motor do you need? The acceleration values are given here, in terms of the gravitational constant $g$, and $F=ma$, so you need to decide on a vehicle mass. But you don't have the mass because you haven't picked the motor, and you can't pick the motor until you have the mass, so now you're into "iterative design," where you guess a vehicle weight, then see what size motor you need, then use that motor mass in your design, and if it's more than your estimated weight then you need to re-run the equations and see what the expected performance is, choose a different motor, etc. etc.
You can avoid some or all of this by being overly conservative in your guesses. If you honestly think that the RC car will weigh 1 kg, then you can use 2 kg for your engineering calculations, find a motor that will do it, and if everything works out to be <2 kg then your vehicle will exceed the design specs.
One of the last items to return to would be that "vehicle top speed" criteria. Looking at Wikipedia for some estimates on rolling friction, it gives "ordinary car tires on concrete" as 0.01 to 0.015. This is used in the equation
$$
F_{\text{friction}} = \mu N \\
$$
where $\mu$ is the friction coefficient (0.01 to 0.015) and $N$ is the "normal force", or force pointing toward the driving surface, which is $N = mg$ for flat surfaces. You can use $N = mg\cos{\theta}$ for the ramp, but I'll refer you back to my earlier comment on being conservative with your estimates so you don't have to keep re-running these calculations. The flat surface will give you the bigger normal force, so use that version for performance specifications and anything that meets that value will exceed it for the ramp.
Now you can look at the power your motor needs to deliver, which is force times speed for linear motion. This means that your motor needs to deliver (conservatively) a force of $0.015 mg$ to overcome rolling friction, while moving at 25 miles per hour, which I'll just convert now to 11.2 meters per second because engineering is much easier in metric.
You can compare the output force requirements, too - the faster acceleration (30 feet) is double the 60-foot acceleration requirement, 0.7 g, the 60-foot acceleration is 0.35 g, climbing the incline is 0.17 g, and maintaining top speed is 0.015 g. This is typical, and why top speed was considered "negligible" at the top.
So now back to the motor power, you need $P = Fv$, or $P = (0.015mg)(11.2 \text{m/s})$. Now you plug in a mass, find the power required, find the force required, work back through the tires and gearbox to find the motor torque required, find a motor that meets or exceeds those specifications, update the estimated weight with the actual motor weight, repeat until your design stabilizes.
For example, if we assume 2.5 kg (about 5.5 pounds), then you find that your mechanical power needs to be (0.0152.59.81*11.2) = 4 watts. If you assume 80 percent efficiency in the drivetrain, then the motor needs to be (4/0.8) = 5 watts. If I google around I find a motor like this (no affiliation or endorsement of the supplier, just a Google result), there's a 5 W motor with a rated speed of 3000 rpm.
3000 rev/min is 50 rev/s, which is 314.15 rad/s, and to get to 25 mph (11.2 m/s), I need $v = r\dot{\theta}$ so $r = v/\dot{\theta}$, so $r = 11.2/314.15$ or wheels with a radius is 3.5 centimeters, or about 1.3 inches.
When I look at tires, I find something like this (again, no affiliation or endorsement), which have an outer diameter of 5.75 inches, or a radius of 2.875 inches. To be able to use those tires, I need to slow the motor by a factor of about (2.875/1.3) = 2.2, while preserving the motor output power. This means I need lower RPM, but higher torque, which is what you get with a gearbox. We'll need a gearbox with a ratio of about 2.2:1, which seems to be in a reasonable range according to this post on an RC forum.
Then you go looking for that transmission, and then you add up the frame, transmission, tires, motor, and batteries, then see if you're above or below 2.5 kg, then keep iterating on designs.
You'll need the drive cycle, or the representative route you plan on taking, in order to determine things like average and peak power draw, which you'll need to specify the power electronics and battery, and then also how long it needs to operate, which you'll need along with the average and peak power draws to specify the battery.
Then you add the batteries and power electronics to the weight and do it all over again.
