If 100 m sprinters accelerate from rest for 3.5 s at 2.8 m/s^2, how far have they run to this point? How long will it take them to complete the 100 m sprint, assuming they maintain their speed the rest of the way?
I found the distance, which is 17 m, and the distance they still have to run to complete the race, which is 83m. I also calculated their second velocity, which is 9.8 m/s. But when I divide 83m by 9.8 m/s,(since that's the velocity for the rest of the race) I don't get the correct answer, which is 12 s.
V = at = 2.8 m/s^2 * 3.5 s = 9.8 m/s,
d1 = 0.5 * 2.8 m/s^2 * (3.5 s)^2 = 17.15 m,
t(total) = 3.5 s + 82.85 m / 9.8 m/s =
3.5 + 8.5 = 12 s.
Thanks!
haha i am doing this question 11 years later :)
To calculate the distance covered during the initial acceleration phase, we can use the formula:
\[ d = \frac{1}{2} \times a \times t^2 \]
where:
- \(d\) is the distance covered,
- \(a\) is the acceleration ($2.8 \, \text{m/s}^2$ in this case),
- \(t\) is the time taken for acceleration (3.5 s in this case).
So, substituting the given values into the formula:
\[ d = \frac{1}{2} \times 2.8 \, \text{m/s}^2 \times (3.5 \, \text{s})^2 \]
Evaluating this expression:
\[ d = \frac{1}{2} \times 2.8 \times 12.25 \, \text{m} \]
\[ d \approx 17 \, \text{m} \]
You've correctly calculated the distance covered during the initial acceleration as 17 m.
Now, to calculate the time taken to complete the 100 m sprint while maintaining the speed (assuming constant speed), we can use the formula:
\[ t = \frac{d}{v} \]
where:
- \(t\) is the time taken to complete the sprint,
- \(d\) is the remaining distance to be covered (83 m in this case),
- \(v\) is the constant velocity (9.8 m/s in this case).
So, substituting the given values into the formula:
\[ t = \frac{83 \, \text{m}}{9.8 \, \text{m/s}} \]
Evaluating this expression:
\[ t \approx 8.47 \, \text{s} \]
The calculated time is approximately 8.47 s, which is different from the given answer of 12 s. It seems there might be an error either in the given answer or in the calculations leading to it. However, I have provided the correct method to calculate the time.