A car traveling at 29.0 m/s (64.9 mi/h) runs into a bridge abutment after the driver falls asleep at the wheel. If the driver is wearing a seat belt and comes to rest within a 1.06-m distance, what is his acceleration (assumed constant)?
Question
A car traveling at 29.0 m/s (64.9 mi/h) runs into a bridge abutment after the driver falls asleep at the wheel. If the driver is wearing a seat belt and comes to rest within a 1.06-m distance, what is his acceleration (assumed constant)?
Solution
To solve this problem, we can use the equation of motion that relates final velocity, initial velocity, acceleration, and distance. The equation is:
v^2 = u^2 + 2as
where: v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
In this case, the final velocity (v) is 0 m/s (since the car comes to rest), the initial velocity (u) is 29.0 m/s, and the distance (s) is 1.06 m.
We can rearrange the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (0 - (29.0)^2) / (2 * 1.06)
a = -29.0^2 / 2.12
a = -400.45 m/s^2
The negative sign indicates that this is a deceleration. So, the driver's acceleration (or rather, deceleration) is 400.45 m/s^2.
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