A rule of thumb for automobile collisions against a rigid barrier is that the collision lasts about .11 seconds, for any initial speed and for any model of automobile. accordingly, the deceleration experienced by an automobile during a collision is directly proportional to the change of velocity (delta v) (with a constant factor of proportionality), and therefore (delta v) can be regarded as a measure of the severity of the collision.

a. if the collision lasts .11 seconds, what is the average deceleration experienced by an automobile in an impact on a rigid barrier at 55 km/hr? 65 km/h? 75km/h?

b. for each of these speeds, what is the crush distance of the front end of the automobile? assume constant deceleration for this calculation.

c. for each of these speeds, what is the average force the seat belt must exert to hold a driver of 75 kg in his seat during the impact.

a. To find the average deceleration experienced by an automobile during a collision, we can use the formula:

deceleration = (change in velocity) / (time)

Given that the collision lasts for 0.11 seconds, we need to convert the initial speeds from km/h to m/s, as the units need to be consistent for the formula.

Converting the speeds to m/s:
55 km/h = (55 * 1000) / (60 * 60) m/s ≈ 15.28 m/s
65 km/h = (65 * 1000) / (60 * 60) m/s ≈ 18.06 m/s
75 km/h = (75 * 1000) / (60 * 60) m/s ≈ 20.83 m/s

Using the formula, we can calculate the average deceleration for each speed:

For 55 km/h:
deceleration = (0 - 15.28) / 0.11 ≈ -139.00 m/s^2

For 65 km/h:
deceleration = (0 - 18.06) / 0.11 ≈ -163.27 m/s^2

For 75 km/h:
deceleration = (0 - 20.83) / 0.11 ≈ -189.36 m/s^2

b. To find the crush distance of the front end of the automobile, we can use the formula:

crush distance = (initial velocity)^2 / (2 * deceleration)

Using the decelerations from part a, we can calculate the crush distances for each speed:

For 55 km/h:
crush distance = (15.28)^2 / (2 * -139.00) ≈ 0.982 m

For 65 km/h:
crush distance = (18.06)^2 / (2 * -163.27) ≈ 0.936 m

For 75 km/h:
crush distance = (20.83)^2 / (2 * -189.36) ≈ 0.876 m

c. To find the average force the seat belt must exert to hold a driver during the impact, we can use Newton's second law of motion:

force = mass * acceleration

Given that the mass of the driver is 75 kg and using the decelerations from part a, we can calculate the average forces for each speed:

For 55 km/h:
force = 75 * 139.00 ≈ 10,425 N

For 65 km/h:
force = 75 * 163.27 ≈ 12,245 N

For 75 km/h:
force = 75 * 189.36 ≈ 14,202 N

Please note that these calculations assume one-dimensional motion and idealized conditions.