Question

stem physics the expression $16t^2$ models the distance in feet that an object falls during the first $t$ seconds after being dropped. what is the distance the object falls during each time? 20. 0.25 second 21. 0.5 second 22. 2 seconds 23. 10 seconds

Explanation:

Problem 20 (0.25 second)

Step1: Substitute t = 0.25

We know the distance formula is \( d = 16t^2 \). Substitute \( t = 0.25 \) into the formula.

Step2: Calculate the value

First, calculate \( t^2=(0.25)^2 = 0.0625 \). Then, multiply by 16: \( 16\times0.0625 = 1 \).

Step1: Substitute t = 0.5

Use the formula \( d = 16t^2 \), substitute \( t = 0.5 \).

Step2: Calculate the value

First, \( t^2=(0.5)^2 = 0.25 \). Then, \( 16\times0.25 = 4 \).

Step1: Substitute t = 2

Substitute \( t = 2 \) into \( d = 16t^2 \).

Step2: Calculate the value

First, \( t^2 = 2^2=4 \). Then, \( 16\times4 = 64 \).

Answer:

The distance is 1 foot.

Problem 21 (0.5 second)