Stopping Distance: Calculate from 108km/h with ½ V x t

[Travian]

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?

Homework Equations

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 I am not sure if its correct.

Are there any mistakes?

Thank You

 

[Herr Malus]

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

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

 

[Travian]

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:/

 

[Herr Malus]

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.

 

[Travian]

90 meters?

 

[Herr Malus]

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

 

[Travian]

Thank you.

By the way, what's wrong with t?

 

[Herr Malus]

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.

 

[Travian]

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