Posted by **Priscilla** on Saturday, December 26, 2009 at 9:03am.

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?

I know you use the equation d = Vit +1/2 at^2 but to find the distance would you just calculate d = vit only.

v = Vo + a t

so

v = 55 - 11 t

when is v = 0 ?

0 = 55 - 11 t

so t = 5 seconds to stop

d = Vo t + (1/2) a t^2

d = 55 (5) - (1/2) 11 (25)

= 275 - 137.5

=137.5 meters

Vo = 110 m/s

- Physics -
**Priscilla**, Saturday, December 26, 2009 at 9:10am
Ok i think i understand. I'm pretty sure you have to use the full equation d= vit +1/2 at^2

- Physics -
**drwls**, Saturday, December 26, 2009 at 9:37am
a) The average velocity when decelerating will be Vo/2 = 27.5 m/s. Multiply that by the time required to stop, Vo/a, to get the stopping distance X.

X = (Vo/a) * (Vo/2)

X = Vo^2/(2 a)= (55)^2/(22) = 137.5 m

Same answer you got, but I used a shortcut

- Physics -
**Priscilla**, Saturday, December 26, 2009 at 9:49am
HAPPY NEW YEAR! Ok, thanks a lot.

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