A 880-kg (1945 lb) dragster, starting from rest, attains a speed of 27.3 m/s (61.0 mph) in 0.60 s.
(a) Find the average acceleration of the dragster during this time interval?
(b) What is the size of the average force on the dragster during this time interval?
(c) Assume the driver has a mass of 66 kg. What horizontal force does the seat exert on the driver?
To find the answers to these questions, we need to use Newton’s second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
Net force = mass × acceleration
(a) To find the average acceleration of the dragster, we can use the formula:
acceleration = (final velocity - initial velocity) / time
Given that the initial velocity is 0 m/s and the final velocity is 27.3 m/s, and the time interval is 0.60 s, we can substitute these values into the formula:
acceleration = (27.3 m/s - 0 m/s) / 0.60 s
acceleration = 27.3 m/s / 0.60 s
acceleration ≈ 45.5 m/s^2
Therefore, the average acceleration of the dragster during this time interval is approximately 45.5 m/s^2.
(b) To find the size of the average force on the dragster during this time interval, we can use Newton’s second law:
force = mass × acceleration
Given that the mass of the dragster is 880 kg and the acceleration is 45.5 m/s^2, we can substitute these values into the formula:
force = 880 kg × 45.5 m/s^2
force ≈ 40040 N
Therefore, the size of the average force on the dragster during this time interval is approximately 40040 Newtons (N).
(c) To find the horizontal force exerted by the seat on the driver, we can consider the net force acting on the driver. The net force is equal to the mass of the driver multiplied by the acceleration:
Net force = mass × acceleration
Given that the mass of the driver is 66 kg and the acceleration is 45.5 m/s^2, we can substitute these values into the formula:
Net force = 66 kg × 45.5 m/s^2
Net force = 3003 N
Therefore, the horizontal force exerted by the seat on the driver is 3003 Newtons (N).