A 1500 m race is divided into 3 equal, 500 m, sections. In the first section, a racer accelerates from rest at 3.0 m/s^2. In section 2, his velocity reduces steadily to 70 m/s and in section 3, he constantly slows until he stops on the finish line. How long did the entire trip take?
stage I:
500 = 3/2 t^2
t = 18.26s
v = 3*18.26 = 54.78m/s
How can he "reduce" his speed from 54 to 70m/s in stage II?
What did I miss? Is there a typo?
the other equation you can use use is:
Vf^2=Vi^2+2AX
sqrt(2(3m/s^2)(500m))=54.78 m/s
same thing.
for the second part you use the same equation and solve for A. Then plug it in to Vf=Vi+AT, solve for T.
for the third you do the same thing.
To find the total time it took for the entire trip, we need to calculate the time taken for each section and then add them together.
Let's start with section 1:
We can use the equation of motion:
v = u + at,
where:
v = final velocity (70 m/s)
u = initial velocity (0 m/s)
a = acceleration (3.0 m/s^2)
t = time taken in section 1 (unknown)
Rearranging the equation, we get:
t = (v - u) / a.
Plugging in the values, we have:
t = (70 m/s - 0 m/s) / 3.0 m/s^2 = 23.33 s.
Therefore, the time taken for section 1 is 23.33 seconds.
Moving on to section 2:
In this section, the velocity reduces steadily from 70 m/s to 0 m/s. Since the acceleration is not specified, we assume it to be uniform throughout section 2.
We can use the equation:
v² = u² + 2as,
where:
v = final velocity (0 m/s)
u = initial velocity (70 m/s)
a = acceleration in section 2 (unknown)
s = distance covered in section 2 (500 m)
Rearranging the equation, we get:
a = (v² - u²) / (2s).
Plugging in the known values, we have:
a = (0² - 70²) / (2 * 500 m) = -0.98 m/s².
For section 2, we have a negative acceleration (-0.98 m/s²) since the velocity is reducing.
Now, we can use the equation of motion:
v = u + at,
where:
v = final velocity (0 m/s)
u = initial velocity (70 m/s)
a = acceleration (-0.98 m/s²)
t = time taken in section 2 (unknown)
Rearranging the equation, we get:
t = (v - u) / a.
Plugging in the values, we have:
t = (0 m/s - 70 m/s) / -0.98 m/s² = 71.43 s.
Therefore, the time taken for section 2 is 71.43 seconds.
Lastly, section 3:
In this section, the racer constantly slows until he stops at the finish line. Since the acceleration is not given, we assume it to be uniform.
Using the equation of motion:
v = u + at,
where:
v = final velocity (0 m/s)
u = initial velocity (unknown)
a = acceleration in section 3 (unknown)
t = time taken in section 3 (unknown)
Since the racer starts with a velocity of 0 m/s in section 3, we can find the initial velocity using the equation u = v - at. Given that the final velocity is 70 m/s and the time taken in section 3 is t, we have:
u = 70 m/s - (0 m/s^2)t = 70 - 0t = 70 - 0 = 70 m/s.
Hence, the initial velocity (u) in section 3 is 70 m/s.
Now, we can use the equation v = u + at, with v = 0 m/s, u = 70 m/s, and a = acceleration in section 3, to find the time taken in section 3:
0 = 70 m/s + a * t.
Since we know that the velocity is reducing and the final velocity is 0 m/s, the acceleration (a) in section 3 will be negative. To simplify the calculation, let's consider the magnitude of the acceleration (|a|).
For section 3, the magnitude of the acceleration is the same as the magnitude of the acceleration in section 1 (3.0 m/s²).
Plugging in the known values, we can rewrite the equation as:
0 = 70 m/s - 3.0 m/s² * t.
Rearranging the equation, we get:
t = 70 m/s / 3.0 m/s².
t ≈ 23.3 s.
Therefore, the time taken for section 3 is approximately 23.3 seconds.
Now, to find the total time for the entire trip, we add the time taken for each section:
Total time = time in section 1 + time in section 2 + time in section 3
Total time = 23.33 s + 71.43 s + 23.3 s = 118.06 s.
Therefore, the entire trip took approximately 118.06 seconds.