A constant force F is applied to a body of mass m that initially is headed east at velocity v0 until its velocity becomes -v0. The total time of travel is 2t. The total distance the body travels in that time is:
A) $\frac{1}{2} \frac{F}{m} t^{2}$ .
B) $\frac{F}{m} t^{2}$ .
C) $v_{0} t-\frac{1}{2} \frac{F}{m} t^{2}$ .
D) $v_{0} t+\frac{1}{2} \frac{F}{m} t^{2}$ .
E) $2\left(v_{0} t+\frac{1}{2} \frac{F}{m} t^{2}\right)$ .

Question

Quizplus · Accepted Answer

The velocity of the body changes by F1F1F1S1F1F1F10v0 over a time interval of t. Therefore, the average speed during this time interval is:
$$\frac{v_1+v_0}{2}=\frac{Ft/m+v_0}{2}$$
The distance travelled during this time interval is:
$$d_1=\frac{1}{2} \left(\frac{Ft}{m}+v_0\right)t=\frac{1}{2}\frac{F}{m}t^2+v_0t$$
Similarly, the velocity changes by F1F1F1S1F1F1F10v0 over the next time interval of t, resulting in an average speed of:
$$\frac{v_2+v_1}{2}=\frac{2Ft/m+v_0}{2}=\frac{Ft/m+v_1}{2}$$
The distance travelled during this time interval is:
$$d_2=\frac{1}{2} \left(\frac{2Ft}{m}+v_0\right)t=\frac{1}{2}\frac{3F}{m}t^2+v_0t$$
Thus, the total distance travelled is:
$$d=d_1+d_2=\frac{1}{2}\frac{4F}{m}t^2+2v_0t$$
Since the total time of travel is 2t, the average speed is:
$$\frac{d}{2t}=\frac{1}{2}\frac{4F}{m}t+2v_0$$
Therefore, the answer is B, 11eb8b15_7518_4162_8952_4f7048256027_TB8845_11.

A Constant Force F Is Applied to a Body of Mass