Question

calculate the first and second velocities of the car with one washer attached to the pulley, using the formulas (v_1 = 0.25 m/t_1), and (v_2 = 0.25 m/(t_2 - t_1)) where (t_1) and (t_2) are the average times the car took to reach the 0.25 and the 0.50 meter marks. record these velocities, rounded to two decimal places, in table e. what is the first velocity of the car with one washer at the 0.25 meter mark? m/s what is the second velocity of the car with one washer at the 0.50 meter mark? m/s

Explanation:

Step1: First - velocity formula application

The formula for the first velocity is $v_1=\frac{0.25}{t_1}$. However, since the value of $t_1$ (the average time to reach the 0.25 - meter mark) is not given, we'll leave it in formula form for now.

Step2: Second - velocity formula application

The formula for the second velocity is $v_2 = \frac{0.25}{t_2 - t_1}$, where $t_2$ is the average time to reach the 0.50 - meter mark and $t_1$ is the average time to reach the 0.25 - meter mark. Without the values of $t_1$ and $t_2$, we can't calculate a numerical answer. But if we assume we have values for $t_1$ and $t_2$, we would substitute them into the formula.

Since we don't have the values of $t_1$ and $t_2$, we can't give a numerical answer. If we assume $t_1 = 1s$ and $t_2=2s$:

Step3: Calculate first velocity (example)

If $t_1 = 1s$, then $v_1=\frac{0.25}{1}=0.25$ m/s.

Step4: Calculate second velocity (example)

If $t_1 = 1s$ and $t_2 = 2s$, then $v_2=\frac{0.25}{2 - 1}=0.25$ m/s.

Answer:

Since $t_1$ and $t_2$ are not given in the problem, we can't provide a definite numerical answer. If we assume $t_1 = 1s$ and $t_2 = 2s$, the first velocity $v_1 = 0.25$ m/s and the second velocity $v_2=0.25$ m/s. In a real - world scenario, you would substitute the actual values of $t_1$ and $t_2$ into the formulas $v_1=\frac{0.25}{t_1}$ and $v_2=\frac{0.25}{t_2 - t_1}$ to get the accurate results.