Questions: The position ( x ) of a cart is described by the equation following expression: x=-4 m+(6 m / s) t-(2 m / s^2) t^2 (2 pt) When does the cart reach the origin (if ever)? EXPLAIN/SHOW WORK!!! (2 pt) Write an equation for the velocity of the cart as a function of time. (2 pt) When does the cart stop and change directions? EXPLAIN your answer

Solution Steps

To find when the cart reaches the origin, set \( x = 0 \) and solve for \( t \): \[ 0 = -4 + 6t - 2t^2 \] Rearrange the equation: \[ 2t^2 - 6t + 4 = 0 \] Solve the quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ t = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 2 \cdot 4}}{2 \cdot 2} \] \[ t = \frac{6 \pm \sqrt{36 - 32}}{4} \] \[ t = \frac{6 \pm 2}{4} \] \[ t = 2 \quad \text{or} \quad t = 1 \] The cart reaches the origin at \( t = 1 \) second and \( t = 2 \) seconds.

The velocity \( v(t) \) is the derivative of the position \( x(t) \): \[ v(t) = \frac{dx}{dt} \] Given \( x(t) = -4 + 6t - 2t^2 \): \[ v(t) = \frac{d}{dt} (-4 + 6t - 2t^2) \] \[ v(t) = 6 - 4t \]

The cart stops when its velocity \( v(t) \) is zero: \[ 6 - 4t = 0 \] Solve for \( t \): \[ 4t = 6 \] \[ t = \frac{6}{4} \] \[ t = 1.5 \, \text{seconds} \] The cart stops and changes direction at \( t = 1.5 \) seconds.

Final Answer