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May 3, 2015

Homework Help: Physics

Posted by Priscilla on Monday, December 28, 2009 at 8:28pm.

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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