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.