A dragster and driver together have mass 929.4 kg. The dragster, starting from rest, attains a speed of 26.5 m/s in 0.55s.
What is the size of the average force on the dragster during this time interval? Answer in units of N.
a=(V-Vo)/t = (26.5-0)/0.55 = 48.2 m/s^2.
F = m*a = 929.4 * 48.2 = 44,780 N.
To find the size of the average force on the dragster, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by the acceleration.
First, let's find the acceleration of the dragster. We can use the equation:
acceleration = change in velocity / time
Given that the initial velocity (u) is 0 m/s and the final velocity (v) is 26.5 m/s, and the time (t) is 0.55s, we can calculate the acceleration:
acceleration = (v - u) / t
acceleration = (26.5 m/s - 0 m/s) / 0.55 s
acceleration = 48.18 m/s^2
Now that we have the acceleration, we can find the force acting on the dragster. Using Newton's second law of motion:
force = mass * acceleration
force = 929.4 kg * 48.18 m/s^2
force ≈ 44815 N
Therefore, the size of the average force on the dragster during this time interval is approximately 44815 N.