Question

2 / 4 points
the position of a particle moving along the x axis is determined by the function $x(t) = bt - ct^2$, where $b = 8.50$ m/s, $c = 4.80$ m/s², and x will be in meters when t is entered in seconds. determine the following.
(a) time when the particle is at the origin (disregard the trivial solution of $t = 0$.)
$\boldsymbol{1.77}$ s
(b) velocity of the particle when it is at the origin (indicate the direction with the sign of your answer.)
$\boldsymbol{-8.50}$ m/s
(c) acceleration of the particle at any time (indicate the direction with the sign of your answer.)
$\boldsymbol{-9.60}$ m/s²
(d) location of the particle when it changes direction
enter a number. ____ m

Explanation:

Step1: Solve for t at origin (x=0)

Set $x(t)=bt-ct^2=0$, factor: $t(b-ct)=0$. Non-trivial solution: $t=\frac{b}{c}=\frac{8.50}{4.80}\approx1.77$ s

Step2: Find velocity at origin

Velocity $v(t)=\frac{dx}{dt}=b-2ct$. Substitute $t=\frac{b}{c}$: $v=b-2c(\frac{b}{c})=b-2b=-b=-8.50$ m/s

Step3: Find acceleration

Acceleration $a(t)=\frac{dv}{dt}=-2c=-2\times4.80=-9.60$ m/s²

Step4: Find direction change position

Direction changes at $v(t)=0$: $b-2ct=0\implies t=\frac{b}{2c}$. Substitute into $x(t)$:
$x= b(\frac{b}{2c}) - c(\frac{b}{2c})^2=\frac{b^2}{2c}-\frac{b^2}{4c}=\frac{b^2}{4c}$
Substitute values: $x=\frac{(8.50)^2}{4\times4.80}=\frac{72.25}{19.20}\approx3.76$ m

Answer:

(a) $1.77$ s
(b) $-8.50$ m/s
(c) $-9.60$ m/s²
(d) $3.76$ m