Suppose that a sprinter can reach maximum speed by accelerating uniformly for 1.7 seconds. What maximum speed must this sprinter maintain in order to equal the world's record of 19.5 seconds for 200 meters?
d = 0.5a*t^2 = 200 m.
0.5a*19.5^2 = 200
190.125a = 200
a = 1.052 m/s^2.
V = a*t = 1.052m/s^2 * 1.7s = 1.788 m/s.
To find the maximum speed that the sprinter must maintain in order to equal the world record time, we can use the formula for uniform acceleration:
\[ V = u + at \]
where \( V \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
In this case, we know that the sprinter reaches maximum speed by accelerating uniformly for 1.7 seconds, so we can assume that the initial velocity \( u \) is 0 since the sprinter starts from rest.
We can rearrange the formula to solve for the final velocity:
\[ V = u + at = 0 + (a \times 1.7) = 1.7a \]
Since the sprinter must cover a distance of 200 meters in 19.5 seconds, we can use the formula for average speed:
\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \]
Rearranging the formula, we can solve for the total distance:
\[ \text{Total Distance} = \text{Average Speed} \times \text{Total Time} = V \times 19.5 \]
We can substitute the expression for the final velocity from the first equation into the second equation:
\[ \text{Total Distance} = (1.7a) \times 19.5 = 33.15a \]
Since the total distance must be 200 meters, we can set up the equation:
\[ 33.15a = 200 \]
Solving for \( a \), we get:
\[ a = \frac{200}{33.15} \approx 6.033 \, \text{m/s}^2 \]
Finally, we can substitute this value of \( a \) into the expression for the final velocity:
\[ V = 1.7a = 1.7 \times 6.033 \approx 10.26 \, \text{m/s} \]
Therefore, the sprinter must maintain a maximum speed of approximately 10.26 m/s in order to equal the world record time of 19.5 seconds for 200 meters.