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 19.5 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 two phases: acceleration and deceleration. For the acceleration phase, since the object starts from rest ($u = 0$), the displacement $s_1$ during acceleration is given by $s_1=ut+\frac{1}{2}a_1t_1^2$. Here $u = 0$, so $s_1=\frac{1}{2}a_1t_1^2$. Let the acceleration be $a_1$ and time be $t_1$.

Step2: Analyze deceleration phase

The deceleration $a_2$ has the same magnitude as $a_1$ (from the symmetry of the acceleration - time graph). The initial velocity for the deceleration phase $u_2$ is the final velocity of the acceleration phase. Using $v = u+a_1t_1$ (where $u = 0$), $v=a_1t_1$. For the deceleration phase, using $v_f=u_2 + a_2t_2$, where $v_f = 0$, $u_2=a_1t_1$ and $a_2=-a_1$, we get $0=a_1t_1 - a_1t_2$, so $t_2=t_1$.

Step3: Calculate displacement during deceleration

The displacement $s_2$ during deceleration is given by $s_2=u_2t_2+\frac{1}{2}a_2t_2^2$. Substituting $u_2=a_1t_1$ and $a_2=-a_1$ and $t_2=t_1$, we have $s_2=a_1t_1\times t_1-\frac{1}{2}a_1t_1^2=\frac{1}{2}a_1t_1^2$. Since $s_1 = 19.5$ m and $s_2=s_1$.

Step4: Calculate total displacement

The total displacement $s=s_1 + s_2$. Since $s_1 = 19.5$ m and $s_2 = 19.5$ m, $s=19.5+19.5 = 39.0$ m.

Answer:

39.0