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
Ok i think i understand. I'm pretty sure you have to use the full equation d= vit +1/2 at^2
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
HAPPY NEW YEAR! Ok, thanks a lot.
a. If the car is going 55 m/s, how many meters will it travel before it stops?
Well, it seems like you've already done the calculations correctly! So, the car will travel 137.5 meters before it comes to a stop. That's enough distance to make your heart race!
b. How many meters will it take to stop a car going twice as fast?
Now, since the initial velocity for this case is 110 m/s, we can use the same equation to find the distance. Plugging in the values, we get:
d = Vit + (1/2)at^2
d = 110(5) + (1/2)(-11)(5)^2
d = 550 + (-137.5)
d = 412.5 meters
So, my speedy friend, it will take 412.5 meters to stop a car that's going twice as fast. Just be glad the brake pedal doesn't have a "hare" setting – it would take even longer to stop!
To find the answer to part a of the question, you would use the equation d = Vit + 1/2at^2. Since the initial velocity (Vi) is given as 55 m/s and the acceleration (a) is -11 m/s^2, you can substitute these values into the equation:
d = (55)(5) + 1/2(-11)(5)^2
= 275 - 137.5
= 137.5 meters
Therefore, the race car will travel 137.5 meters before it stops.
Now, let's move on to part b of the question. You are given that the initial velocity (Vi) is 110 m/s, which is twice as fast as the previous scenario. To find the distance it takes for the car to stop in this case, we will follow the same steps:
First, we need to find the time it takes for the car to stop. We use the equation v = Vi + at, where the final velocity (v) is 0 m/s:
0 = 110 - 11t
Solving for t, we get:
11t = 110
t = 10 seconds
Using the equation d = Vit + 1/2at^2, we can now find the distance (d):
d = (110)(10) + 1/2(-11)(10)^2
= 1100 - 550
= 550 meters
Therefore, it will take 550 meters for the car to stop when going twice as fast.