A car traveling 91km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.93m .
What was the magnitude of the average acceleration of the driver during the collision?
Express the answer in terms of "g's," where 1.00g=9.80m/s2
To find the magnitude of the average acceleration of the driver during the collision, we can use the following formula:
Acceleration (a) = (vf - vi) / t
Where:
- vf is the final velocity
- vi is the initial velocity
- t is the time
In this case, we are given the initial velocity as 91 km/h, but we need to convert it to m/s because the unit of acceleration is m/s^2. To convert km/h to m/s, we divide by 3.6:
vi = (91 km/h) / (3.6 m/s) = 25.28 m/s
Next, we are given the displacement, which is the distance the driver traveled before coming to a stop. We need to convert it from meters to kilometers so that the units align with the initial velocity:
s = 0.93 m = 0.93 m / 1000 = 0.00093 km
Now, we need to find the time it took for the driver to come to a stop. We'll use the formula of motion:
s = (vi + vf) * t / 2
Since the car is initially traveling at 25.28 m/s and it comes to rest, vf = 0 m/s.
0.00093 km = (25.28 m/s + 0 m/s) * t / 2
Simplifying the equation, we get:
0.00093 km = (25.28 m/s) * t / 2
0.00093 km = 12.64 m/s * t
0.00093 km = (12.64 m/s) * t
To solve for t, we can rearrange the equation:
t = (0.00093 km) / (12.64 m/s) = 0.0000736 hours
Now, we can substitute the values of vi = 25.28 m/s and t = 0.0000736 hours into the acceleration formula:
Acceleration (a) = (vf - vi) / t
Acceleration (a) = (0 m/s - 25.28 m/s) / 0.0000736 hours
Converting hours to seconds:
0.0000736 hours = 0.0736 seconds
Acceleration (a) = (0 m/s - 25.28 m/s) / 0.0736 s
Acceleration (a) = -25.28 m/s / 0.0736 s
Acceleration (a) = -343.5 m/s^2
Finally, we can express the answer in terms of "g's" by dividing the acceleration by the value of 1g:
Acceleration (a) = (-343.5 m/s^2) / (9.80 m/s^2)
Acceleration (a) = -35.08 g's
Therefore, the magnitude of the average acceleration of the driver during the collision is approximately 35.08 g's in the opposite direction of the initial velocity.