Question

the graph represents the motion of an object over an interval of time in a straight line, starting from rest at time t = 0. the horizontal axis shown represents zero acceleration. the object travels 11.5 meters while speeding up. find the total distance (in meters) that the object has traveled at the moment it comes to rest. provide at least one decimal place

Explanation:

Step1: Recall kinematic - equations

The motion has three phases: acceleration, constant - velocity (not shown in the problem but conceptually there), and deceleration. For the acceleration phase, using the kinematic equation $v^{2}=v_{0}^{2}+2a\Delta x$. Since $v_{0} = 0$ (starts from rest), $v^{2}=2a\Delta x$. The distance during acceleration $\Delta x_{1}=11.5$ m.

Step2: Analyze deceleration phase

The magnitude of acceleration during deceleration is the same as during acceleration (from the symmetry of the problem, as the acceleration - time graph has equal - magnitude non - zero accelerations). Using the kinematic equation $v_{f}^{2}=v^{2}+2(-a)\Delta x_{2}$. When the object comes to rest, $v_{f} = 0$. So, $0 = v^{2}-2a\Delta x_{2}$, which implies $\Delta x_{2}=\Delta x_{1}$ (because $v^{2}=2a\Delta x_{1}$ from Step 1).

Step3: Calculate total distance

The total distance $d=\Delta x_{1}+\Delta x_{2}$. Since $\Delta x_{1}=\Delta x_{2}=11.5$ m, $d = 2\times11.5=23.0$ m.

Answer:

23.0