In getting ready to slam-dunk the ball, a basketball player starts from rest and sprints to a speed of 11.0 m/s in 2.58 s. Assuming that the player accelerates uniformly, determine the distance he runs.
v = 0 + a t
11 = a (2.58)
a = 11/2.58
d = (1/2) a t^2
= (1/2)(11/2.58) (2.58)^2
or simply
=the average speed = 5.5
5.5 * 2.58 will give the same answer :)
thanks Demon
Well, let's break it down, shall we? First, we need to find the acceleration. We can do that by using the equation:
acceleration = change in velocity / time
So, the change in velocity is 11.0 m/s (since the player goes from rest to 11.0 m/s) and the time is 2.58 seconds. Therefore, the acceleration is:
acceleration = 11.0 m/s / 2.58 s
Now, to determine the distance the player runs, we can use another equation:
distance = initial velocity * time + (1/2) * acceleration * time^2
Since the player starts from rest, the initial velocity is 0. Plugging in the values we have, we get:
distance = 0 * 2.58 + (1/2) * (11.0 m/s / 2.58 s) * (2.58 s)^2
distance = 0 + (1/2) * 4.26 m/s^2 * 6.6564 s^2
distance = 0 + 2.13 m/s^2 * 6.6564 s^2
distance = 14.23 meters
So, the basketball player runs a distance of approximately 14.23 meters.
To determine the distance the player runs, we can use the equation of motion:
\[ v = u + at \]
Where:
- \( v \) is the final velocity (given as 11.0 m/s)
- \( u \) is the initial velocity (which is 0 m/s since the player starts from rest)
- \( a \) is the acceleration (uniform since it is mentioned in the question)
- \( t \) is the time taken to reach the final velocity (given as 2.58 s)
We can rearrange the equation to solve for acceleration:
\[ a = \frac{{v - u}}{{t}} \]
Substituting the given values:
\[ a = \frac{{11.0 \, \text{m/s} - 0 \, \text{m/s}}}{{2.58 \, \text{s}}} \]
Simplifying:
\[ a = \frac{{11.0 \, \text{m/s}}}{{2.58 \, \text{s}}} \]
Now, we can use another equation of motion to find the distance:
\[ s = ut + \frac{1}{2}at^2 \]
Where:
- \( s \) is the distance
- \( u \) is the initial velocity (0 m/s)
- \( t \) is the time taken to reach the final velocity (2.58 s)
- \( a \) is the acceleration (\( \frac{{11.0 \, \text{m/s}}}{{2.58 \, \text{s}}} \))
Substituting the given values:
\[ s = 0 \times 2.58 + \frac{1}{2} \left( \frac{{11.0 \, \text{m/s}}}{{2.58 \, \text{s}}} \right) \times \left(2.58 \, \text{s} \right)^2 \]
Simplifying:
\[ s = \frac{1}{2} \times 11.0 \, \text{m/s} \times 2.58 \, \text{s} \]
\[ s = \frac{1}{2} \times 28.38 \, \text{m} \]
\[ s = 14.19 \, \text{m} \]
Therefore, the distance the player runs is 14.19 meters.