A car going 15 m/s is brought to rest in a distance of 3 m as it strikes a pile of dirt. How large a force is exerted by the seatbelts on a 90 kg passenger as the car is stopped?
first, find the acceleration needed.
Then, F = ma
To determine the force exerted by the seatbelts on the passenger, we can use Newton's second law of motion, which states that force (F) is equal to the mass (m) of an object multiplied by its acceleration (a). Here, the car is initially moving at a velocity of 15 m/s and is brought to rest over a distance of 3 m.
First, let's determine the acceleration. We can use the following kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the car comes to rest)
u = initial velocity (15 m/s)
a = acceleration
s = distance (3 m)
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the values, we have:
a = (0^2 - 15^2) / (2 * 3)
a = -225 / 6
a = -37.5 m/s² (negative because the car is decelerating)
Now that we have the acceleration, we can calculate the force exerted on the passenger using Newton's second law:
F = m * a
Substituting the mass (m) of the passenger as 90 kg and the acceleration (a) as -37.5 m/s², we get:
F = 90 kg * (-37.5 m/s²)
F = -3,375 N
The negative sign indicates that the force exerted by the seatbelts on the passenger is in the opposite direction of the motion.
Therefore, the force exerted by the seatbelts on the 90 kg passenger as the car is stopped is -3,375 Newtons.