An 18-wheeler with a mass of 35,000 kg is moving at a speed of 65 mph or about 30 m/s. If this vehicle collides with a barrier and the collision lasts one tenth of a second (0.1 s), what is the force experienced by this vehicle?
To calculate the force experienced by the vehicle, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a), or F = m*a.
First, let's calculate the acceleration of the vehicle during the collision. We can use the equation a = Δv / Δt, where Δv is the change in velocity and Δt is the change in time.
The change in velocity (Δv) can be calculated as the final velocity (v2) minus the initial velocity (v1), or Δv = v2 - v1.
Given that the initial velocity (v1) is 30 m/s and the final velocity (v2) is 0 m/s (as the vehicle comes to a stop during the collision), we have Δv = 0 m/s - 30 m/s = -30 m/s (the negative sign indicates a deceleration).
Now, we can calculate the acceleration (a) using Δv and the change in time (Δt).
Given that Δt = 0.1 s, we have a = Δv / Δt = -30 m/s / 0.1 s = -300 m/s^2.
Since acceleration is a vector quantity and points in the opposite direction of the initial velocity, the negative sign indicates that the acceleration is in the opposite direction to the motion of the vehicle.
Now, we can calculate the force experienced by the vehicle during the collision.
F = m*a = 35,000 kg * -300 m/s^2 = -10,500,000 N.
The force experienced by the vehicle during the collision is -10,500,000 Newtons.