Coasting due west on your bicycle at 7.9 , you encounter a sandy patch of road 7.0 across. When you leave the sandy patch your speed has been reduced by 2.0 to 5.9 .
1. Assuming the sand causes a constant acceleration, what was the bicycle's acceleration in the sandy patch? Give the magnitude.
2.How long did it take to cross the sandy patch?
3. Suppose you enter the sandy patch with a speed of only 5.4 . Is your final speed in this case 3.4 , more than 3.4 , or less than 3.4 ? Explain.
1. 1.7
2. 1.2
3. <3.4
To solve these problems, we can use the equations of motion in the x-direction. Let's denote the initial speed as v0, the final speed as vf, the distance as d, and the acceleration as a.
1. From the given information, we have:
Initial speed, v0 = 7.9 m/s
Final speed, vf = 5.9 m/s
Distance, d = 7.0 m
From the equation of motion vf^2 = v0^2 + 2ad, we can solve for the acceleration a:
(5.9 m/s)^2 = (7.9 m/s)^2 + 2a(7.0 m)
34.81 m^2/s^2 = 62.41 m^2/s^2 + 14a
-27.60 m^2/s^2 = 14a
a = -27.60 m^2/s^2 / 14
a ≈ -1.97 m/s^2
Therefore, the bicycle's acceleration in the sandy patch is approximately 1.97 m/s^2.
2. To find the time it takes to cross the sandy patch, we can use the equation of motion vf = v0 + at and solve for t:
5.9 m/s = 7.9 m/s + (-1.97 m/s^2)t
-1.97 m/s^2t = -2.0 m/s
t = (-2.0 m/s) / (-1.97 m/s^2)
t ≈ 1.02 s
So, it takes approximately 1.02 seconds to cross the sandy patch.
3. If the initial speed is 5.4 m/s, and the sandy patch still causes the same constant acceleration, we can use the same equation vf = v0 + at to find the final speed:
vf = 5.4 m/s + (-1.97 m/s^2)t
Since the time to cross the sandy patch (t) remains the same, the final speed will be:
vf = 5.4 m/s + (-1.97 m/s^2)(1.02 s)
vf ≈ 3.34 m/s
Therefore, the final speed in this case is approximately 3.34 m/s, which is less than 3.4 m/s.
1. To find the bicycle's acceleration in the sandy patch, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = ma). In this case, the force is caused by the sand, and we can assume the mass of the bicycle remains constant. Therefore, we can rearrange the equation to solve for acceleration:
acceleration = force / mass
We know that the force acting on the bicycle is responsible for the change in its speed. So, using the equation:
force = mass x change in velocity
The change in velocity is the difference between the initial speed (7.9 m/s) and the final speed (5.9 m/s).
change in velocity = 7.9 m/s - 5.9 m/s = 2.0 m/s
Given that the width of the sandy patch is 7.0 m, we can now calculate the acceleration using the formula:
acceleration = (change in velocity) / time
However, we don't have the time yet, so let's move on to question 2 to find out.
2. To calculate the time it took to cross the sandy patch, we can use the formula:
distance = speed x time
We already know the width of the sandy patch, which is 7.0 m, and the final speed after leaving the sandy patch, which is 5.9 m/s. Substituting these values into the formula, we get:
7.0 m = 5.9 m/s x time
Solving for time, we divide both sides of the equation by 5.9 m/s:
time = 7.0 m / 5.9 m/s ≈ 1.19 s
Now that we have the time, we can go back to question 1 to calculate the acceleration:
acceleration = (change in velocity) / time
= 2.0 m/s / 1.19 s
≈ 1.68 m/s²
Therefore, the bicycle's acceleration in the sandy patch is approximately 1.68 m/s².
3. If you enter the sandy patch with a speed of only 5.4 m/s, your final speed will be less than 3.4 m/s. This can be explained by the fact that the sandy patch causes a deceleration or negative acceleration, meaning it works to slow down the bicycle. As observed in question 1, the bicycle decelerated from an initial speed of 7.9 m/s to a final speed of 5.9 m/s. Since your initial speed of 5.4 m/s is already lower than 7.9 m/s, it will experience an even greater decrease in speed in the sandy patch, resulting in a final speed lower than 5.9 m/s.