The national transportation safety bureau indicates that a person in a car crash has a reasonable chance of survival if his or her acceleration is less than 300 m/s^2. suppose that the mass of a car is 1200kg . what stopping distance is needed if the initial speed before the collision is 23m/s ?

We can use constant acceleration equation to solve for the distance.

Formula is:

Vf^2 = Vi^2 + 2ad

where Vf^2 is final velocity squared, Vi^2 is initial velocity squared, a is acceleration (or deceleration) and d is the distance.

we want the car to come to complete stop, that is, Vf^2 be equal to zero.

Therefore, the equation becomes 0 = Vi^2 + 2ad. Solving for d we get:

d = (-(Vi)^2)/2a). We can ignore the minus sign since acceleration is really deceleration.

We know initial velocity (23m/s) and we know acceleration (max= 300 m/s^2). Plugging these in, we get:

d = ((23 m/s)^2)/(2* 300m/s^2) = 0.88m

hope that helps

real, smaller and upside-down.   so the answer should be b.

answeralert drivers who drift off the road;