Question

a baseball travels ( d ) meters ( t ) seconds after being dropped from the top of a building. the distance traveled by the baseball can be modeled by the equation ( d = 5t^2 ).

1. complete the table and plot the data on the coordinate plane.

( t ) (seconds) | ( d ) (meters)
--- | ---
0 |
0.5 |
1 |
1.5 |
2 |

(there is a coordinate plane with x - axis labeled distance traveled (meters) from 0 to 30 and y - axis labeled time (seconds) from 0 to 3.5)

1. is the baseball traveling at a constant speed? explain how you know.

Explanation:

1. Completing the Table and Plotting Data

We are given the equation \( d = 5t^2 \), where \( d \) is the distance in meters and \( t \) is the time in seconds. We need to find the distance \( d \) for different values of \( t \) (0, 0.5, 1, 1.5, 2 seconds) by substituting each \( t \) into the equation.

Step 1: For \( t = 0 \) seconds

Substitute \( t = 0 \) into \( d = 5t^2 \):
\( d = 5(0)^2 = 0 \) meters.

Step 2: For \( t = 0.5 \) seconds

Substitute \( t = 0.5 \) into \( d = 5t^2 \):
\( d = 5(0.5)^2 = 5(0.25) = 1.25 \) meters.

Step 3: For \( t = 1 \) second

Substitute \( t = 1 \) into \( d = 5t^2 \):
\( d = 5(1)^2 = 5(1) = 5 \) meters.

Step 4: For \( t = 1.5 \) seconds

Substitute \( t = 1.5 \) into \( d = 5t^2 \):
\( d = 5(1.5)^2 = 5(2.25) = 11.25 \) meters.

Step 5: For \( t = 2 \) seconds

Substitute \( t = 2 \) into \( d = 5t^2 \):
\( d = 5(2)^2 = 5(4) = 20 \) meters.

Now, we can complete the table:

\( t \) (seconds)	0	0.5	1	1.5	2

To plot the data, we use the coordinate plane where the x - axis represents time \( t \) (in seconds) and the y - axis represents distance \( d \) (in meters). We plot the points \((0,0)\), \((0.5,1.25)\), \((1,5)\), \((1.5,11.25)\), and \((2,20)\) on the coordinate plane.

2. Determining if the Baseball is Traveling at a Constant Speed

A constant speed means that the distance traveled per unit time (the speed) is the same. Speed \( v \) is calculated as \( v=\frac{\Delta d}{\Delta t}\), where \(\Delta d\) is the change in distance and \(\Delta t\) is the change in time. We will calculate the speed for different time intervals and check if it is constant.

Step 1: Calculate speed between \( t = 0 \) and \( t = 0.5 \)

\(\Delta t=0.5 - 0 = 0.5\) seconds, \(\Delta d = 1.25-0 = 1.25\) meters.
Speed \( v_1=\frac{\Delta d}{\Delta t}=\frac{1.25}{0.5}=2.5\) m/s.

Step 2: Calculate speed between \( t = 0.5 \) and \( t = 1 \)

\(\Delta t = 1 - 0.5=0.5\) seconds, \(\Delta d=5 - 1.25 = 3.75\) meters.
Speed \( v_2=\frac{\Delta d}{\Delta t}=\frac{3.75}{0.5} = 7.5\) m/s.

Since \( v_1 = 2.5\) m/s and \( v_2=7.5\) m/s are not equal, the speed is not constant. Another way to see this is from the equation \( d = 5t^2 \). The equation is a quadratic equation (not linear), and for motion with constant speed, the distance - time equation should be linear (\( d=vt \), where \( v \) is constant). Since \( d \) is a quadratic function of \( t \), the speed (which is the derivative of \( d \) with respect to \( t \), \( v=\frac{dd}{dt}=10t\)) is a function of time and increases as time increases.

Answer:

s:

1. The completed table is:

\( t \) (seconds)	0	0.5	1	1.5	2

(For plotting, plot the points \((0,0)\), \((0.5,1.25)\), \((1,5)\), \((1.5,11.25)\), \((2,20)\) on the given coordinate plane with time on the x - axis and distance on the y - axis.)

1. No, the baseball is not traveling at a constant speed. We know this because the speed (calculated as the rate of change of distance with respect to time) changes as time increases (e.g., speed between \( t = 0 \) and \( t = 0.5 \) is 2.5 m/s and between \( t = 0.5 \) and \( t = 1 \) is 7.5 m/s) or because the distance - time relationship is quadratic (\( d = 5t^2 \)) rather than linear (which would be the case for constant speed, \( d=vt \)).