contestada

A box proceeds along the x-axis and the figure below shows a record of its velocity as a function of time. Every grid line along the vertical axis corresponds to 2.00 m/s and each gridline along the horizontal axis corresponds to 0.500 s. (Enter your answer in m/s^2. Indicate the direction with signs of your answers. Note that t=0 at the intersection of the axes.) a) Determine the average acceleration of the box in the time interval t=0 to t=2.50 s. b) Determine the average acceleration of the box in the time interval t=2.50 s to t=7.50 s. c) Determine the average acceleration of the box in the time interval t=0 to t=10.0 s

Respuesta :

Answer:

The answer is "[tex]\bold{0, 12.8 \ \frac{m}{s^2}, and \ 6.4 \ \frac{m}{s^2}}[/tex]"

Explanation:

For point a:

The average time interval acceleration for the box [tex]t = 0\ to\ t = 250 \ s:[/tex]

[tex]\to \alpha_{0 \ to \ 2.50}=\frac{v_f - v_i}{t_f-t_i}=\frac{0-0}{2.50-0}=0[/tex]

For point b:

The average time interval acceleration for the box

[tex]t = 250 \ s\ to\ t = 750 \ s :[/tex]  

[tex]\to \alpha_{2.50 \ to \ 7.50}=\frac{v_f - v_i}{t_f-t_i}=\frac{32-(-32)}{7.50-2.50}=12.8 \ \frac{m}{s^2}[/tex]

For point c:

The average time interval acceleration for the box [tex]t = 0\ to \ t = 10 \ s :[/tex]  

[tex]\to \alpha_{2.50 \ to \ 7.50}=\frac{v_f - v_i}{t_f-t_i}=\frac{32-(-32)}{10-0}=6.4 \ \frac{m}{s^2}[/tex]