In a head-on collision, a car stops in 0.24 from a speed of 15 . The driver has a mass of 77 , and is, fortunately, tightly strapped into his seat.
What force is applied to the driver by the seatbelt in that fraction of a second?
Well, you know what they say - seatbelts are like hugs from the car! So, in this particular scenario, the force applied to the driver by the seatbelt can be calculated using Newton's second law, which states that force (F) equals mass (m) times acceleration (a).
Given that the car stops in 0.24 seconds, we can calculate the acceleration using the formula a = Δv / t, where Δv is the change in velocity. Initially, the car is moving at a speed of 15 m/s and comes to a stop, so Δv = 15 m/s - 0 m/s = -15 m/s.
Now, let's find the acceleration: a = (-15 m/s) / (0.24 s) = -62.5 m/s² (negative because the car is decelerating).
Next, we can calculate the force applied to the driver using F = m * a, where m is the mass of the driver. The question states that the mass of the driver is 77 kg, so let's crunch the numbers: F = 77 kg * -62.5 m/s² = -4812.5 N.
So, the force applied to the driver by the seatbelt in that fraction of a second is approximately -4812.5 N, or should we say the force of a really intense car-hugging session! Stay safe on those roads!