Question

Starting from rest at home plate, a baseball player runs to first base (90 ft away). He uniformly accelerates over the first 10.9 ft to his maximum speed, which is then maintained until he crosses first base. If the overall run is completed in 4.5 seconds, determine his maximum speed, the acceleration over the first 10.9 feet, and the time duration of the acceleration.

Answer 1

Answer:

• Maximum speed $ 20.16 \frac{ft}{s}$
• The acceleration is $37.33 \frac{ft}{s^2}$
• The time duration of the acceleration is $ 0.5401 s$

Explanation:

We will using the following kinematics equations for position and speed at time t:

$x(t) = \ x_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2$

$v(t) \ = \ v_0 \ + \ a \ t$

where $x_0$ is the initial position, $v_0$ the initial speed, and a the acceleration.

Lets take this problem in two parts, part 1 and part 2, of duration $t_1$ and $t_2$ respectively.

first part

For the first part, we know that the player starts from rest, this is, the initial speed is zero, and accelerates uniformly. So, we can use the following equations:

The final speed will be:

$v(t_1) \ = \ v_0 \ + \ a \ t_1$

$v_{max} = a \ t_1$

and the distance traveled

$x(t_1) = \ x_0 \ + \ v_0 \ t_1 \ + \frac{1}{2} \ a \ t_1^2$

$x(t_1) =  \frac{1}{2} \ a \ t_1^2 = 10.9 ft$

so

$a \ t_1^2 = 10.9 ft$

second part

Now, we know that at maximum speed, the player run a distance $90 ft - 10.9 ft$, so:

$v_{max} * t_2 = 90 ft - 10.9 ft = 79.8 ft$

All together

Taking the equations

$v_{max} = a \ t_1$

$a \ t_1^2 = 10.9 ft$

$v_{max} * t_2 =  79.8 ft$

and

$t_1 + t_2 = 4.5 s$

we can take the second equation and make

$a \ t_1^2 = (a \ t_1 )* t_1 =  10.9 ft$

but looking at the first equation, this is

$v_{max} t_1 = 10.9 ft$

now, adding the third one

$v_{max} t_1  + v_{max} * t_2 =   10.9 ft + 79.8 ft$

$v_{max} (t_1  + t_2) =   90.78 ft$

using the fourth

$v_{max} 4.5 s =   90.78 ft$

$v_{max} =   \frac{90.78 ft}{ 4.5 s}$

$v_{max} =  20.16 \frac{ft}{s}$

so, this is the maximum speed.

Now, if this is the speed, then, $t_1$ is...

$v_{max} t_1 = 20.16 \frac{ft}{s} t_1 = 10.9 ft$

$ t_1 = \frac{10.9 ft}{20.16 \frac{ft}{s}}$

$ t_1 = 0.5401 s$

and this is the time duration of the acceleration

Finally, using

$v_{max} = a \ t_1$

we found

$a = \frac{v_{max}}{t_1}$

$a = \frac{20.16 \frac{ft}{s}}{ 0.5401 s}$

$a = 37.33 \frac{ft}{s^2}$