An object begins at position x = 0 and moves one-dimensionally along the x-axis witļi a velocity v
expressed as a function of time t according to the graph above. At what time does the object pass
through x = 0 again?

Respuesta :

Answer:

The answer is "between 20 s and 30 s".

Explanation:

Calculating the value of positive displacement:

[tex]\ (x_{+ve}) = \frac{1}{2} \times 15 \times 20 \\\\[/tex]

          [tex]= \frac{1}{2} \times 300 \\\\= 150 \\\\[/tex]

Calculating the value of negative displacement upon the time t:

[tex](x_{-ve}) = \frac{1}{2} \times 5 \times 20- 20(t-20) \\\\[/tex]

          [tex]= \frac{1}{2} \times 100- 20t+ 400 \\\\= 50- 20t+ 400 \\\\[/tex]

[tex]\to X= X_{+ve} + X_{-ve} \\\\[/tex]

[tex]\to 150 - 50 -20t+400 =0\\\\\to 100 -20t+400 =0 \\\\\to 500 -20t =0\\\\\to 20t =500 \\\\\to t=\frac{500}{20}\\\\\to t=\frac{50}{2}\\\\\to t= 25[/tex]

That's why its value lie in "between 20 s and 30 s".