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Ksenya-84 [330]
4 years ago
7

The National Transportation Safety Bureau indicates that a person in a car crash has a reasonable chance of survival if his or h

er 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 ?
Physics
1 answer:
Aleks [24]4 years ago
4 0

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) = <span>0.88m </span>

<span>hope that helps</span>

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In this problem, the object we are analyzing is the coffee cup.

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8 0
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