A car traveling at 72 km/hr hits a tree. the driver, who has a mass of 55.0 kg comes to a rest 0.10s after the impact.
What is the force that acts on the driver?
72 km/hr = 20 m/s
So, the deceleration is (20m/s)/(.1s) = 200m/s^2
Now the force F=ma, so ...
I would solve it this way:
force*time=mass*changeinvelocity
8980
To calculate the force that acts on the driver, we need to use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
First, let's convert the speed of the car from km/hr to m/s to match the unit of time (seconds) we'll be using in the calculations.
1 km/hr = 1000 m / 3600 s = 0.2778 m/s
So, the speed of the car is 72 km/hr × 0.2778 m/s = 20 m/s.
Since the driver comes to rest after the impact, we can assume the driver experiences uniform deceleration during that time.
To calculate the acceleration, we can use the formula v = u + at, where:
v = final velocity (0 m/s in this case, as the driver comes to rest)
u = initial velocity (20 m/s in this case, as the driver was traveling at that speed)
a = acceleration
t = time (0.10 s in this case)
0 = 20 m/s + a × 0.10 s
Rearranging the equation:
a × 0.10 s = -20 m/s
a = (-20 m/s) / (0.10 s) = -200 m/s²
The negative sign indicates that the acceleration is in the opposite direction to the car's initial motion.
Now that we have the acceleration, we can calculate the force using Newton's second law:
Force = mass × acceleration
The mass of the driver is given as 55.0 kg:
Force = 55.0 kg × (-200 m/s²)
Force = -11,000 N
The force acting on the driver is -11,000 N, with the negative sign indicating that it acts in the opposite direction to the car's initial motion.