An 879-kg (1943 lb) dragster, starting from rest, attains a speed of 25.6 m/s (57.3 mph) in 0.59 s.

(a) Find the average acceleration of the dragster during this time interval.

(b) What is the magnitude of the average net force on the dragster during this time?

I am not going to do all these for you. They are all basically the same.

a = (V2-V1)/t = (25.6-0)/.59 = 43.4 m/s^2

F = m a = 879 * 43.4 = 38,140 Newtons

To find the average acceleration of the dragster, we can use the formula:

acceleration = (final velocity - initial velocity) / time

Given:
- Initial velocity (v₀) = 0 m/s (starting from rest)
- Final velocity (v) = 25.6 m/s
- Time (t) = 0.59 s

(a) Find the average acceleration of the dragster:

acceleration = (25.6 m/s - 0 m/s) / 0.59 s
acceleration = 25.6 m/s / 0.59 s
acceleration ≈ 43.4 m/s²

So, the average acceleration of the dragster during this time interval is approximately 43.4 m/s².

Now, to find the magnitude of the average net force on the dragster during this time, we can use Newton's second law of motion:

Net force = mass * acceleration

Given:
- Mass (m) = 879 kg

(b) What is the magnitude of the average net force on the dragster during this time?

Net force = 879 kg * 43.4 m/s²
Net force ≈ 38,123 N

So, the magnitude of the average net force on the dragster during this time is approximately 38,123 Newtons.