You are driving through town at 12.0 m/s when suddenly a ball rolls out in front of you. You apply the brakes and begin decelerating at 3.5 m/s^2.(a)How far do you travel before stopping?(b)When you have traveled only half the distance in part(a), is your speed 6.0 m/s, greater than 6.0 m/s, or less than 6.0 m/s? support your answer with a calcation.
x=20.57m then
(b) V= 8.48m/s , so it's greater ..use the V^2= Vo^2 +2a(X2-Xo)
To solve this problem, we have the initial velocity (Vi) as 12.0 m/s, the acceleration (a) as -3.5 m/s^2, and we need to find the distance traveled before stopping. We can use the kinematic equation:
v^2 = Vi^2 + 2aΔx
For part (a), we need to find the distance traveled before stopping, so we need to find Δx. At the stopping point, v = 0 m/s. Plugging in the given values:
0 = (12.0 m/s)^2 + 2(-3.5 m/s^2)Δx
Rearranging the equation, we get:
168 = -7.0 Δx
Dividing both sides of the equation by -7.0:
Δx = -24.0 m
Since distance cannot be negative, we take the absolute value:
Δx = 24.0 m
Therefore, the car travels 24.0 meters before stopping.
For part (b), we need to determine the speed when the car has traveled half the distance found in part (a). So, the distance traveled is 24.0 m / 2 = 12.0 m.
Using the formula v^2 = Vi^2 + 2aΔx, plugging in the given values:
v^2 = (12.0 m/s)^2 + 2(-3.5 m/s^2)(12.0 m)
Calculating:
v^2 = 144 m^2/s^2 + (-84.0 m^2/s^2)
v^2 = 60.0 m^2/s^2
Taking the square root of both sides:
v = √(60.0 m^2/s^2)
v ≈ 7.75 m/s
Therefore, when the car has traveled half the distance, the speed is approximately 7.75 m/s. Hence, it is greater than 6.0 m/s.
To answer part (a) of the question, we need to find the distance traveled before coming to a stop. We can use the kinematic equation:
vf^2 = vi^2 + 2aΔx
Where:
vf = final velocity (which is 0 m/s in this case, since we are coming to a stop)
vi = initial velocity (12.0 m/s)
a = acceleration (-3.5 m/s^2)
Δx = distance traveled
Plugging in the given values, we have:
0^2 = (12.0 m/s)^2 + 2(-3.5 m/s^2)(Δx)
Simplifying:
0 = 144 m^2/s^2 - 7.0 m/s^2(Δx)
Rearranging the equation to solve for Δx:
7.0 m/s^2(Δx) = 144 m^2/s^2
Δx = 144 m^2/s^2 / 7.0 m/s^2
Δx = 20.57 m
So, the distance you travel before coming to a stop is approximately 20.57 meters.
Moving on to part (b) of the question, we need to determine the speed when we have traveled only half the distance found in part (a).
First, we divide the distance traveled in part (a) by 2:
20.57 m / 2 = 10.28 m
Now, we use the following equation of motion to find the final speed at this distance:
vf^2 = vi^2 + 2aΔx
Where:
vf = final velocity (which is what we are trying to find)
vi = initial velocity (12.0 m/s)
a = acceleration (-3.5 m/s^2)
Δx = 10.28 m
Plugging in the values:
vf^2 = (12.0 m/s)^2 + 2(-3.5 m/s^2)(10.28 m)
Simplifying:
vf^2 = 144 m^2/s^2 - 2(3.5 m/s^2)(10.28 m)
vf^2 = 144 m^2/s^2 - 71.68 m^2/s^2
vf^2 = 72.32 m^2/s^2
Taking the square root of both sides to solve for vf:
vf = √(72.32 m^2/s^2)
vf ≈ 8.5 m/s
So, when you have traveled only half the distance found in part (a), your speed is approximately 8.5 m/s.