(a) What is the acceleration of a car in m/s² that doubles its speed from 36.0 km/h in 5.00 s?

(b) What net force acted on this car if its mass was 1,000.0 kg?

a. V = 36000m/h / 3600s/h = 10 m/s.

a = (Vf-Vo) / t = (20-10) / 5=2 m/s^2

Fn = ma = 1000 * 2 = 2000 N.

To find the acceleration of a car that doubles its speed, we first need to convert the given initial speed from km/h to m/s, and then determine the change in velocity.

(a) Conversion from km/h to m/s:
1 km/h = 1000 m/3600 s (converting distance to meters and time to seconds)
= 1000/3600 m/s
≈ 0.278 m/s

So, the initial speed of the car is 36.0 km/h × 0.278 m/s ≈ 10.0 m/s.

The change in velocity is given by the final velocity minus the initial velocity. Since the final velocity is twice the initial velocity:

Change in velocity = 2 × initial velocity = 2 × 10.0 m/s = 20.0 m/s

To find the acceleration, we can use the formula:

acceleration = change in velocity / time

The time given is 5.00 s. Substituting the values into the formula, we get:

acceleration = 20.0 m/s / 5.00 s = 4.00 m/s²

Therefore, the acceleration of the car is 4.00 m/s².

(b) To find the net force acting on the car, we can use Newton's second law of motion:

force = mass × acceleration

The mass of the car is given as 1,000.0 kg. Substituting this value and the previously calculated acceleration into the formula, we get:

force = 1,000.0 kg × 4.00 m/s² = 4,000.0 kg·m/s²

Hence, the net force acting on the car is 4,000.0 kg·m/s², which can also be expressed as 4,000.0 N (Newtons).