Question

In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acceleration of 50g that lasts for less than 30 ms, but in a crash with a 50g acceleration lasting longer than 30 ms, a driver is unlikely to survive. Imagine a collision in which a driver's head experienced a 50g acceleration. What is the shortest survivable distance over which the driver's head could have come to rest?

Answer 1

Answer:

0.22 m

Explanation:

We are told that the driver can survive an acceleration of 50g only if the collision lasts no longer than 30 ms. So,

$t = 30 ms = 0.030 s$

The acceleration is

$a=-50g = -50(9.8)=-490 m/s^2$

where the negative sign is due to the fact that this is a deceleration, since the driver comes to a stop in the collision.

First of all, we can find what the initial velocity of the car should be in this conditions by using the equation:

$v=u+at$

And since the final velocity is zero, v=0, and solving for u,

$u=-at=-490(0.030)=14.7 m/s$

And now we can find the corresponding distance travelled using the equation:

$d=ut+\frac{1}{2}at^2 = (14.7)(0.030)+\frac{1}{2}(-490)(0.030)^2=0.22 m$