A set of crash tests consists of running a test car moving at a speed of 12.2 m/s (27.3 mi/hr) into a solid wall. Strapped securely in an advanced seat belt system, a 75.0 kg (165 lbs) dummy is found to move a distance of 0.630 m from the moment the car touches the wall to the time the car is stopped. Calculate the size of the average force which acts on the dummy during that time.
Using the direction of motion as the positive direction, calculate the average acceleration of the dummy during that time (in g's, with 1g = 9.81m/s2).In a different car, the distance the dummy moves while being stopped is reduced from 0.630 m to 0.270 m,
calculate the average force on the dummy as that car stops.
To calculate the average force acting on the dummy during the crash test, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
First, let's calculate the acceleration of the dummy during the crash test using the formula:
a = (vf - vi) / t
where vf is the final velocity, vi is the initial velocity, and t is the time taken.
Given:
Initial velocity (vi) = 12.2 m/s
Final velocity (vf) = 0 m/s (since the car is stopped)
Time taken (t) = Unknown
To find the time taken, we need to rearrange the equation:
t = (vf - vi) / a
We can assume that the dummy moves uniformly during the crash test, so the time taken for the dummy to move from the moment the car touches the wall to the time the car is stopped is the same as the time taken for the car to stop.
Substituting the given values, we get:
t = (0 - 12.2) / a
Now, we need to calculate the acceleration in terms of g's. Since 1g = 9.81 m/s^2, we divide the acceleration value by 9.81 to convert it into g's.
a = 0.630 m / t (Equation 1)
Similarly, for the second car where the distance the dummy moves is reduced to 0.270 m, we can use:
a' = 0.270 m / t (Equation 2)
To find the average force acting on the dummy, we can substitute the calculated acceleration values into the equation F = m * a.
Let's solve Equation 1 for the first car:
a = (0.630 m) / ((0 - 12.2) / a)
a^2 = (0.630 m) * a / (0 - 12.2)
a^2 + (0.630 m / 12.2) * a = 0
Solving this quadratic equation for a, we get:
a ≈ -16.177 g's or 16.177 g's
Since we're interested in the average acceleration, we take the positive value: a ≈ 16.177 g's.
Now, calculating the average force for the first car using F = m * a:
F = (75.0 kg) * (16.177 * 9.81 m/s^2)
F ≈ 11,920.53 N
Thus, the average force acting on the dummy during the crash test in the first car is approximately 11,920.53 N.
For the second car, we use Equation 2 to calculate the average acceleration:
a' = (0.270 m) / ((0 - 12.2) / a')
a'^2 = (0.270 m) * a' / (0 - 12.2)
a'^2 + (0.270 m / 12.2) * a' = 0
Solving this quadratic equation for a', we get:
a' ≈ -5.805 g's or 5.805 g's
Taking the positive value, a' ≈ 5.805 g's.
Calculating the average force for the second car using F = m * a':
F = (75.0 kg) * (5.805 * 9.81 m/s^2)
F ≈ 4,267 N
Thus, the average force acting on the dummy during the crash test in the second car is approximately 4,267 N.