Question

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?

Answer 1

Answer:

(a) 0 s

(b) 4.00 m

(c) -0.76 s

(d) +0.76 s

Step-by-step explanation:

$x=4.00-7.00t^2$

It stops momentarily when the velocity, $v$ is 0. $v$ is the derivate of $x$.

(a) $v=\frac{dx}{dt}=-14.00t$

Setting this to 0,

$-14.00t=0$

$t=0$

(b) Substitute this value for $t$ in $x$ to get its position.

$x=4.00-7.00\times0^2=4.00$ m

It passes the origin when $x=0$

$4.00-7.00t^2=0$

$7.00t^2=4$

$t^2=\frac{4}{7}$

$t=\pm\sqrt{\frac{4}{7}}$

(c) The negative time is $t=-\sqrt{\frac{4}{7}} =-0.76 s$

(d) The positive time is $t=+\sqrt{\frac{4}{7}} =+0.76 s$