Question

Question #7f631

Answer 1

`-"33.4g"`

Explanation:

Start by converting the speed of the car from kilometers per hour to meters per second

`85 color(red)(cancel(color(black)("km")))/color(red)(cancel(color(black)("h"))) * (10^3"m")/(1color(red)(cancel(color(black)("km")))) * (1color(red)(cancel(color(black)("h"))))/"3600 s" = "23.61 m s"^(-1)`

Now, notice that the car goes from a speed of `"23.61 m s"^(-1)` to rest, i.e. to a speed of `"0 m s"^(-1)`, so right from the start, you should expect the average acceleration to come out negative.

In other words, the car is decelerating, which means that its acceleration is acting in the opposite direction to its direction of movement.

Another thing to notice here is that the speed of the car is decreasing significantly over a very short distance. From the moment of impact, it takes only

`d = "0.85 m"`

for the speed of the car to decrease by `"23.61 m s"^(-1)`, This tells you that the acceleration of the car will have a very high absolute value, i.e. its magnitude will be very high.

Your tool of choice here will be this equation

`color(blue)(ul(color(black)(v_f^2 = v_0^2 + 2 * a * d)))`

Here

• `v_f` is the final speed of the car
• `v_0` is its initial speed, i.e. its speed before the impact
• `a` is its acceleration
• `d` is the distance covered by the driver after the impact

Rearrange the above equation to solve for `a`

`2 * a * d = v_f^2 - v_0^2 implies a = (v_f^2 - v_0^2)/(2 * d)`

Plug in your values to find

`a = ( 0^2 "m"^color(red)(cancel(color(black)(2))) "s"^(-2) - 23.61^2 "m"^color(red)(cancel(color(black)(2))) "s"^(-2))/(2 * 0.85 color(red)(cancel(color(black)("m")))) = -"327.9 m s"^(-2)`

As you can see, the acceleration is indeed negative. Convert the acceleration to `"g"`'s to get

`-327.9 color(red)(cancel(color(black)("m s"^(-2)))) * "1.00g"/(9.80color(red)(cancel(color(black)("m s"^(-2))))) = color(darkgreen)(ul(color(black)(-"33.4g")))`

I'll leave the answer rounded to three sig figs.