A race car can be slowed with a constant acceleration of -11 m/s^2

a. If the car is going 55 m/s, how many meters will it travel before it stops?

b. How many meters will it take to stop a car going twice as fast?

According to one of the teachers this is what they stated

A. stopping distance = (stopping time) x (average speed)
= (V/a)*(V/2) = V^2/(2a)

B. If the speed doubles, with deceleration rate "a" staying the same, the stopping distance is four times farther.

Ok, to find the distance for the first one i understand. I'm pretty sure this is how you do it

d = Vit
d = 55 x 5
d = 275 m

But for b, Would i have to double the speed for this which is 55 x 2 = 110 and would i also have to increase the stoppping distance which is 275 m by multiplying by 4

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  1. This question has been answered twice already.

    Yes, the stopping distance quadruples if you double the speed.

    What is there about
    Stopping distance = V^2/(2a)
    that you don't understand?

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  2. Ok, i just understand this a little. I know it has been answered twice by you and i thank you for that. Can you tell me if this right below?

    a. stoppping distance = v^2 / 2a
    sd = 55^2 / 2(- 11)
    sd = 3025 / - 22
    sd = -137.5 m

    b. Therefore, if the stopping distance quadruples then i shouldn't put that in the equation of:

    sd = v^2 / 2a
    sd = 110^2 / 2 (- 11)
    sd = 12100 / -22
    sd = -550 m

    Is this right? Thanks i appreciate this a lot and happy new yr.

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