# Stopping distance

## Homework Statement

The maximum straight-line deceleration of a racing car under braking is 5 m s-2. What is the minimum stopping distance of the car from a velocity of 108 km h-1?

½ V x t

## The Attempt at a Solution

V = 108km/h=30m/s

t = 30/ 5 = 6 m s-2

½ V x t = ½ x 30 x 6 = 90

Answer: Distance = 90 m

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I found this formula ½ V x t (to determine stopping distance) on the internet, but im not sure if its correct.

Are there any mistakes?

Thank You

## Answers and Replies

It seems like your formula for distance is the average distance travelled in a given time for an object moving at constant speed. I would Instead try the formula:
vf 2 = vo 2 + 2 a $$\Delta$$ x

Note that you are decelerating so the plus in front of the 2 is actually a minus here.

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so: vf = 30 m/s -2 x 5 m/s...
what is this delta x? sorry, haven't slept this night at all, feeling a bit dizzy now:/

Sorry, delta x here stands for the change in position of the car.

Basically, how far it goes in stopping.

Edit: Sorry I'm without a calculator so I hadn't noticed but this gives the same answer you already had.

90 meters?

Yes, 90 meters. Aside from the units in your 't', everything seems to be in order.

Thank you.

By the way, whats wrong with t?

Units: I'm a little mixed up by your scheme but it looks like you have t (time) in meters per second per second, instead of just seconds.

Oh seriously... I fixed that. Thanks again:)