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?
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
2 answers
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.
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.
1 answer