An Olympic athlete wants to complete the 4000-m run in less than 12.0 minutes. After exactly 10.0 minutes of running at a constant speed, she still has 800 m to go. If she then accelerates at a rate of 0.40 m/ s 2 , a) how much longer will it take her to complete the race? b) will she achieve her desired time?
To find the time it will take her to complete the race, we need to find the distance she needs to cover while accelerating. Subtracting the distance she already covered from the total distance gives us: 4000m - 800m = 3200m.
We can now use the equation of motion:
vf^2 = vi^2 + 2*a*d
where vf is the final velocity, vi is the initial velocity (zero since she started from rest), a is the acceleration, and d is the distance.
Rearranging the equation to solve for vf, we have:
vf = sqrt(2*a*d)
vf = sqrt(2 * 0.40 m/s^2 * 3200 m)
vf ≈ 32 m/s
So, her final velocity is approximately 32 m/s.
To find the time it will take her to reach this final velocity, we use the equation:
vf = vi + a*t
32 m/s = 0 + (0.40 m/s^2)(t)
32 m/s = 0.40 m/s^2 * t
t = 32 m/s / 0.40 m/s^2
t ≈ 80 s
Therefore, it will take her an additional 80 seconds to complete the race.
The total time it will take her to complete the race is: 10 minutes + 80 seconds = 10 minutes and 1 minute and 20 seconds.
Converting this to minutes, we have:
10 minutes + 1.33 minutes = 11.33 minutes
Thus, it will take her approximately 11.33 minutes to complete the race.
Since her desired time is less than 12.0 minutes, she will achieve her desired time.
A falling flowerpot takes 0.20 s to fall past a window that is 1.9 m tall. From what height above the top of the window was the flower- pot dropped?
We can use the equations of motion to solve this problem. Let's assume the initial height of the flowerpot is h meters above the top of the window.
The equation we can use is:
h = vi*t + (1/2)*a*t^2
In this case, the initial velocity vi is 0 since the flowerpot starts from rest, and the acceleration a is due to gravity, which is approximately 9.8 m/s^2.
Plugging in the values we know:
1.9 m = 0 * 0.20 s + (1/2)*9.8 m/s^2 * (0.20 s)^2
Simplifying the equation:
1.9 m = 0 + (1/2)*9.8 m/s^2 * 0.04 s^2
1.9 m = 0.0196 m
To find the initial height h, we subtract this value from the height of the window:
h = 1.9 m - 0.0196 m = 1.8804 m
Therefore, the flowerpot was dropped from a height of approximately 1.8804 meters above the top of the window.
A hot-air balloon is rising upward with a con- stant velocity of 4.0 m/ s. As the balloon reaches a height of 4.0 m above the ground, the balloonist accidentally drops a can of pop over the edge of the basket. How long does it take the pop can to reach the ground?
To find the time it takes for the pop can to reach the ground, we can use the equation of motion:
d = vi * t + (1/2) * a * t^2
In this case, the pop can is dropped from rest, so the initial velocity vi is 0. The acceleration a is due to gravity, which is approximately -9.8 m/s^2 (negative because it acts downwards).
The distance d is the height at which the pop can was dropped, which is 4.0 m.
Plugging in the known values into the equation, we have:
4.0 m = 0 * t + (1/2) * (-9.8 m/s^2) * t^2
Simplifying the equation:
4.0 m = (-4.9 m/s^2) * t^2
To solve for t, we can rearrange the equation:
t^2 = (4.0 m) / (-4.9 m/s^2)
t^2 ≈ -0.8163 s^2
Since time cannot be negative, we can conclude that the pop can will not reach the ground.
Therefore, the pop can will not reach the ground and the time it takes to reach the ground is undefined.
A stone is dropped off a cliff of height h. At the same time, a second stone is thrown straight upward from the base of the cliff with an initial velocity v �� i . Assuming that the second rock is thrown hard enough, at what time t will the two stones meet?
To find the time at which the two stones meet, we need to set up equations for the respective motions of the two stones and find when their positions are equal.
Let's denote h as the height of the cliff, vi as the initial velocity of the second stone, and t as the time at which the stones meet.
For the stone dropped from the cliff:
The equation of motion is:
h_cliff = (1/2) * g * t^2
Since the stone is dropped from rest, the initial velocity is 0, and g is the acceleration due to gravity, approximately -9.8 m/s^2.
For the stone thrown upwards:
The equation of motion is:
h_throw = vi * t + (1/2) * g * t^2
To find the time t when the two stones meet, we set h_cliff equal to h_throw:
(1/2) * g * t^2 = vi * t + (1/2) * g * t^2
Simplifying the equation:
t^2 * g / 2 = vi * t
t * (t * g / 2 - vi) = 0
Since the time t cannot be zero, we can solve for t when the expression in the parenthesis is equal to zero:
t * g / 2 - vi = 0
t * g / 2 = vi
t = (2 * vi) / g
Therefore, the time at which the two stones meet is given by the equation (2 * vi) / g.