In a head-on collision, a car stops in 0.14{\rm s} from a speed of 24{\rm m}/{\rm s} . The driver has a mass of 67{\rm kg} , and is, fortunately, tightly strapped into his seat.
What force is applied to the driver by his seat belt during that fraction of a second?
11486N. Find velocity 24/0.14= 171.4m/s^2 Then F=MA 171.4*67=11485.71
In a head-on collision, a car stops in 0.13s from a speed of 26m/s . The driver has a mass of 62kg , and is, fortunately, tightly strapped into his seat. What force is applied to the driver by his seat belt during that fraction of a second?
To find the force applied to the driver by his seat belt during the collision, we can use Newton's second law of motion, which states that:
force = mass x acceleration
First, we need to find the acceleration of the driver. We can use the kinematic equation:
final velocity = initial velocity + acceleration x time
Given that the initial velocity is 24 m/s and the time is 0.14 s, we can rearrange the equation to solve for acceleration:
acceleration = (final velocity - initial velocity) / time
acceleration = (0 - 24) m/s / 0.14 s
acceleration = -171.4 m/s^2
The negative sign indicates that the acceleration is in the opposite direction of the initial velocity.
Next, we can calculate the force applied to the driver by the seat belt using Newton's second law:
force = mass x acceleration
Given that the driver's mass is 67 kg, we can substitute the values into the equation:
force = 67 kg x (-171.4 m/s^2)
force = -11492 N
Therefore, the force applied to the driver by his seat belt during that fraction of a second is -11492 Newtons. The negative sign indicates that the force is in the opposite direction of the initial velocity.
To find the force applied to the driver by his seat belt during the collision, we can apply Newton's second law of motion, which states that the force (F) acting on an object is equal to the object's mass (m) multiplied by its acceleration (a). In this case, the acceleration can be found using the given final velocity (0 m/s), initial velocity (24 m/s), and time (0.14 s).
First, let's find the acceleration:
The formula to find acceleration is
a = (vf - vi) / t
Where:
a = acceleration
vf = final velocity
vi = initial velocity
t = time
Substituting the given values, we have:
a = (0 - 24) / 0.14
a = -24 / 0.14
a = -171.43 m/s^2 (Note: The negative sign indicates the deceleration of the car)
Now that we have the acceleration, we can find the force applied to the driver by his seat belt:
F = m * a
Where:
F = force
m = mass
a = acceleration
Substituting the given mass, we get:
F = 67 kg * (-171.43 m/s^2)
F = -11,477.81 N
The force applied to the driver by his seat belt during the collision is approximately -11,477.81 Newtons. The negative sign indicates that the force is in the opposite direction of the initial motion. This force is necessary to decelerate the driver and prevent them from moving forward due to inertia.