In running the 40 m dash in tryouts fot the Notre Dame football team, Zach S. completes the 40 m in 4.8 seconds. Assuming constant acceleration throughout the run, find his velocity at the end of 40 m and his acceleration during the 40 m.
You know, if you have a problem with time, acceleration, velocity and distance, before bothering to post the question, try using the formulas you know and love:
v = at
s = 1/2 at^2
Since there is constant acceleration,
s = 1/2 at^2
40 = 1/2 a * 4.8^2
a = 3.47222 m/s^2
v = at = 3.47222 * 4.8 = 16.67 m/s
To find Zach's velocity at the end of the 40 m and his acceleration during the run, we can use the equations of motion.
The first equation we can use is the equation that relates displacement, initial velocity, final velocity, and acceleration:
\(v^2 = u^2 + 2as\)
Where:
- \( v \) is the final velocity
- \( u \) is the initial velocity
- \( a \) is the acceleration
- \( s \) is the displacement
In this case, we know that the displacement \( s \) is 40 meters, and the initial velocity \( u \) is 0 (since Zach starts from rest). We want to find the final velocity \( v \) and the acceleration \( a \).
Plugging in the known values into the equation, we have:
\( v^2 = 0^2 + 2a(40) \)
\( v^2 = 80a \)
Now, since we know that Zach completes the 40 m sprint in 4.8 seconds, we can use another equation of motion that relates displacement, initial velocity, time, and acceleration:
\( s = ut + \frac{1}{2}at^2 \)
Plugging in the known values, we have:
\( 40 = 0(4.8) + \frac{1}{2}a(4.8)^2 \)
\( 40 = 11.52a \)
Now we have a system of two equations with two variables. We can solve these equations simultaneously to find the values of \( v \) and \( a \).
Multiplying the first equation by 11.52, we get:
\( 11.52v^2 = 921.6a \)
We can substitute this value of \( 921.6a \) into the second equation:
\( 40 = 11.52a \)
By rearranging this equation, we find the value of \( a \):
\( a = \frac{40}{11.52} \)
\( a \approx 3.47 \, \text{m/s}^2 \)
Now, substituting this value of \( a \) into the first equation, we can find the value of \( v \):
\( v^2 = 80(3.47) \)
\( v^2 \approx 277.6 \)
Taking the square root of both sides, we find:
\( v \approx 16.67 \, \text{m/s} \)
Therefore, Zach's velocity at the end of the 40 m is approximately 16.67 m/s and his acceleration during the run is approximately 3.47 m/s^2.
For the acceleration a, solve
X = (1/2) a t^2
a = 2X/t^2 = 3.472 m/s^2
For the velocity after 4.8 seconds, solve
V = a t
You will get a number that is higher than the world record average speed for the 100 m dash. I do not believe the figures. The problem is probably the assumption that constant acceleration can be maintained for 40 meters.