A runner hopes to complete the 10000-m run in less than 30min. After running at constant speed for exactly 27min, there are still 1200m to go. The runner must then accelerate at 0.20m/s^2 for how many seconds in order to achieve the desired time?
Here is a solution from eight years ago by Mathmate. Note in his problem the distance remaining was 1100 meters, so adjust that for your 1200.
<< A runner hopes to complete the 10,000m run in less than 30 minutes. After exactly 27 minutes, there are still 1100m to go. The runner must then accelerate at 0.20 m/s^2 for how many seconds in order to achieve the desired time?
I get 80.9 but I know that the answer is 3.1 seconds. How do you get 3.1 seconds? (What formula/processes?) Thanks.
Physics Problem - MathMate Monday, August 31, 2009 at 1:04am
Assuming he ran at uniform speed for the first 27 minutes. Then initial speed,
u = (10000-1100)m/(27*60)sec. = 5.494 m/s
Remaining time is 3 minutes, or 180 s.
Distance to run, S = 1100m
a = acceleration = 0.2 m/s
let t=time of acceleration, then
new speed=u+at
time to run at new speed=(180-t) sec.
Distance run during acceleration
=ut+(1/2)at²
Summing up distance run in 3 minutes,
1100 = ut+(1/2)at² + (u+at)*(180-t)
which simplifies to
t² -360t +10000/9 = 0
From which t can be solved using the quadratic formula as
t=3.113 sec. or t=356.887 sec.
We reject the second solution because it exceeds the 180 sec. limit.
To calculate the time required to accelerate, we need to use the following equation:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the initial velocity (u) is unknown, the final velocity (v) is the velocity required to complete the remaining 1200m in the desired time, and the acceleration (a) is given as 0.20m/s^2.
We can use the equation of motion to find the initial velocity (u) before acceleration:
s = ut + (1/2)at^2
Where:
s = total distance remaining
u = initial velocity
a = acceleration
t = time
In this case, since we know the remaining distance (s) is 1200m, the acceleration (a) is 0.20m/s^2, and the time (t) is unknown. We can rearrange the equation to solve for the initial velocity (u):
u = (2s - at^2)/(2t)
Substituting the values, we get:
u = (2 * 1200 - 0.20 * t^2)/(2 * t)
Now, we can substitute this value of initial velocity (u) into the first equation mentioned to find the time (t) required for acceleration:
v = u + at
Substituting the values, we get:
v = (2 * 1200 - 0.20 * t^2)/(2 * t) + 0.20 * t
Since the desired time is 30 minutes, we need to convert it to seconds (since acceleration is given in m/s^2):
30 minutes * 60 seconds/minute = 1800 seconds
Let's simplify the equation:
v = (2400 - 0.20t^2)/(2t) + 0.20t
Now, let's solve for t. Rearrange the equation to isolate the term with t:
v = (2400 - 0.20t^2)/(2t) + 0.20t
Multiply both sides by 2t:
2tv = 2400 - 0.20t^2 + 0.40t^2
Combine the like terms:
0.20t^2 + 2tv = 2400
Rearrange the equation:
0.20t^2 + 2tv - 2400 = 0
Now, we have a quadratic equation that can be solved using the quadratic formula. The quadratic equation is in the form: ax^2 + bx + c = 0, where:
a = 0.20
b = 2v
c = -2400
The quadratic formula is:
t = (-b ± √(b^2 - 4ac))/(2a)
Substitute the values:
t = (-(2v) ± √((2v)^2 - 4 * 0.20 * -2400))/(2 * 0.20)
t = (-2v ± √(4v^2 + 1920))/(0.40)
From the problem statement, we know that after 27 minutes of constant speed, there are still 1200m to go. So, the remaining distance is s = 1200m.
To find v, we use the formula:
v = s/t
Substituting the values, we get:
v = 1200m/(30min - 27min) = 1200m/3min = 400m/min
Now, substitute this value of v into the equation for t:
t = (-2 * 400 ± √((4 * 400^2 + 1920))/(0.40)
Now, you can solve the equation for t using a calculator.
To find out how many seconds the runner must accelerate, we can start by finding the initial speed and the remaining distance after 27 minutes.
Given:
Total time = 30 minutes = 1800 seconds
Time elapsed = 27 minutes = 1620 seconds
Remaining distance = 1200 meters
First, let's calculate the initial speed of the runner:
Initial speed = Distance / Time
= Remaining distance / Remaining time
= 1200m / (1800s - 1620s)
= 1200m / 180s
= 6.67 m/s
Now that we have the initial speed, we can find the final speed using the acceleration equation:
Final speed = Initial speed + (Acceleration x Time)
Let's assume the runner accelerates for t seconds.
Final speed = Initial speed + (0.20 m/s^2) x t
Now, we need to find the time it takes for the runner to cover the remaining distance at the final speed.
Remaining distance = (Final speed x Time) + (0.5 x Acceleration x Time^2)
= (Final speed x t) + (0.5 x 0.20 m/s^2 x t^2)
Since the remaining distance is 1200m, we can substitute the values and solve for t:
1200m = (Final speed x t) + (0.1 m/s^2 x t^2)
Now, let's substitute the value of the final speed we found earlier:
1200m = (6.67 m/s + (0.20 m/s^2 x t)) x t + (0.1 m/s^2 x t^2)
Next, we can simplify the equation:
1200m = 6.67m/s x t + (0.20 m/s^2 x t^2) + (0.1 m/s^2 x t^2)
1200m = (6.67 m/s + 0.30 m/s^2) x t^2
Now, we can isolate t^2:
t^2 = 1200m / (6.97 m/s)^2
t^2 = 24.35 s^2
Taking the square root of both sides:
t = √(24.35 s^2)
t ≈ 4.94 seconds
Therefore, the runner must accelerate for approximately 4.94 seconds in order to achieve the desired time.