Acceleration is sometimes expressed in multiples of g, where is the acceleration of an object due to the earth's gravity. in a car crash, the car's forward velocity may go from to in how many g's are experienced, on average, by the driver?

Driver experiences 17.68 G's on an average.

In the question,

We know that in a car crash the speed of the car goes from the speed of 26 m/s to 0 m/s in 0.15 seconds.

So,

The final speed, v = 0 m/s

Initial speed, u = 26 m/s

Also,

Time in seconds, t = 0.15 s

So,

Using the equation of law of the motion, we get,

v = u + at

0 = 26 + a(0.15)

a = -26/0.15

a = -173.33 m/s²

As, there is deceleration of the car is taking place the acceleration is in negative.

Now,

To convert it into the force in the G's we need to divide the deceleration by the acceleration (gravity), g.

So,

G's experienced by the driver on an average is given by,

Therefore, on an average the driver experiences 17.68 G's.

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.