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?
average speed during stop = 7.5 m/s
t = 3 meters/7.5 m/s = .4 second
v = 15 + a t
0 = 15 + .4 a
a = -37.5 m/s^2
F = ma = 90(-37.5) = -3375 Newtons
ouch !
To find the force exerted by the seatbelts on the passenger, we can use the equation for stopping distance and deceleration.
Given:
Initial velocity, u = 15 m/s
Stopping distance, s = 3 m
Mass of the passenger, m = 90 kg
To calculate the deceleration, we can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (which is 0 since the car comes to rest)
u = initial velocity
a = deceleration
s = stopping distance
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Plugging in the values:
a = (0^2 - 15^2) / (2 * 3)
a = (-225) / 6
a = -37.5 m/s^2 (negative sign indicates deceleration)
Now, to find the force exerted by the seatbelts, we can use Newton's second law of motion:
F = ma
Where:
F = force
m = mass of the passenger
a = deceleration
Plugging in the values:
F = 90 kg * (-37.5 m/s^2)
F = -3375 N
The force exerted by the seatbelts on the passenger is -3375 Newtons. The negative sign indicates that the force acts in the opposite direction to the motion of the car.
To determine the force exerted by the seatbelts on the passenger, we can use the concept of deceleration and Newton's second law of motion.
Newton's second law states that the force acting on an object is equal to its mass multiplied by its acceleration. In this case, the car is brought to rest, so its final velocity is 0 m/s. The initial velocity of the car is given as 15 m/s.
First, let's calculate the deceleration (negative acceleration) of the car. We can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s)
u = initial velocity (15 m/s)
a = acceleration (deceleration, since the car is slowing down)
s = distance (3 m)
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Substituting the values:
a = (0 - 15^2) / (2 * 3)
a = -225 / 6
a = -37.5 m/s^2
Now that we have the deceleration, we can calculate the force exerted on the passenger using Newton's second law:
F = ma
Substituting the values:
F = 90 kg * -37.5 m/s^2
F = -3375 N
Therefore, the force exerted by the seatbelts on the 90 kg passenger as the car is stopped is 3375 Newtons (N). Note that the negative sign indicates that the force is exerted in the opposite direction of the car's motion, which is necessary to bring the car to rest.