A car travelling at 81 km/h slams on the brakes and comes to a stop after 2.9 s. What is the force from the seatbelt acting on the driver if the driver has a mass of 72 kg?
To find the force from the seatbelt acting on the driver, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration, or F = m * a.
First, we need to calculate the acceleration of the car. We can use the formula for acceleration, which is a = (vf - vi) / t, where vf is the final velocity, vi is the initial velocity, and t is the time taken.
Given that the car comes to a stop, the final velocity (vf) will be 0 km/h. The initial velocity (vi) is given as 81 km/h. However, we need to convert these velocities to m/s to ensure consistent units.
Converting 0 km/h to m/s:
vf = 0 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 0 m/s
Converting 81 km/h to m/s:
vi = 81 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 22.5 m/s
Now, we can calculate the acceleration:
a = (vf - vi) / t
a = (0 m/s - 22.5 m/s) / 2.9 s
a = -22.5 m/s / 2.9 s
a = -7.76 m/s^2
Since the car is decelerating, the acceleration is negative.
Next, we can calculate the force applied to the driver using Newton's second law of motion:
F = m * a
F = 72 kg * (-7.76 m/s^2)
F ≈ -559.5 N
Therefore, the force from the seatbelt acting on the driver is approximately 559.5 N. The negative sign indicates that the force is acting in the opposite direction to the car's motion, which is the force of deceleration.
To determine the force from the seatbelt acting on the driver, we can use Newton's second law:
F = ma
where F is the force, m is the mass, and a is the acceleration.
First, we need to find the deceleration of the car. We can use the following kinematic equation:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
In this case, initial velocity (vi) is 81 km/h, final velocity (vf) is 0 (since the car comes to a stop), and the time (t) is 2.9 s.
Converting the initial velocity to meters per second:
vi = (81 km/h) * (1000 m/1 km) * (1/3600 h/1 s) = 22.5 m/s
Using the kinematic equation, we can solve for the deceleration (a):
0 = 22.5 m/s + a * 2.9 s
Rearranging the equation:
a = -22.5 / 2.9 ≈ -7.76 m/s²
Since acceleration is the rate of change of velocity, but in the opposite direction, we take its magnitude.
Now, we can use Newton's second law to find the force:
F = m * a
Substituting in the values:
F = (72 kg) * (-7.76 m/s²) ≈ -559.68 N
The negative sign indicates that the force is in the opposite direction of the motion, which is necessary to bring the car to a stop. Therefore, the force from the seatbelt acting on the driver is approximately 559.68 N.