x axis is x = 4.00 - 7.00t2, with x in meters and t in seconds. (a) At what time and (b) where does the particle (momentarily) stop? At what (c) negative time and (d) positive time does the particle pass through the origin?

Respuesta :

Answer:

(a) 0 s

(b) 4.00 m

(c) -0.76 s

(d) +0.76 s

Step-by-step explanation:

[tex]x=4.00-7.00t^2[/tex]

It stops momentarily when the velocity, [tex]v[/tex] is 0. [tex]v[/tex] is the derivate of [tex]x[/tex].

(a) [tex]v=\frac{dx}{dt}=-14.00t[/tex]

Setting this to 0,

[tex]-14.00t=0[/tex]

[tex]t=0[/tex]

(b) Substitute this value for [tex]t[/tex] in [tex]x[/tex] to get its position.

[tex]x=4.00-7.00\times0^2=4.00[/tex] m

It passes the origin when [tex]x=0[/tex]

[tex]4.00-7.00t^2=0[/tex]

[tex]7.00t^2=4[/tex]

[tex]t^2=\frac{4}{7}[/tex]

[tex]t=\pm\sqrt{\frac{4}{7}}[/tex]

(c) The negative time is [tex]t=-\sqrt{\frac{4}{7}} =-0.76 s[/tex]

(d) The positive time is [tex]t=+\sqrt{\frac{4}{7}} =+0.76 s[/tex]