Uniform accelerated motion (UAM) equations:

$\overline{)\mathbf{}{{\mathit{v}}}_{{\mathit{f}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathbf{}}{\mathbf{+}}{\mathit{a}}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{x}}{\mathbf{=}}{\mathbf{}}\mathbf{\left(}\frac{{\mathit{v}}_{\mathit{f}}\mathbf{+}{\mathit{v}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\right)}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{x}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathit{t}}{\mathbf{+}}{\frac{1}{2}}{\mathit{a}}{{\mathit{t}}}^{{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{{{\mathit{v}}}_{{\mathit{f}}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{}}{{{\mathit{v}}}_{{\mathbf{0}}}}^{{\mathbf{2}}}{\mathbf{}}{\mathbf{+}}{\mathbf{2}}{\mathit{a}}{\mathbf{\u2206}}{\mathit{x}}}$

Assume 1 horizontal block is 1s.

$\mathbf{\u2206}\mathit{x}\mathbf{=}\mathbf{}\mathbf{\left(}\frac{{\mathit{v}}_{\mathit{f}}\mathbf{+}{\mathit{v}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\right)}\mathit{t}$

At A:

- v
_{f,1}= 0.5 - v
_{0}_{,1}= 0 - v
_{f,2}= -1.75 - v
_{0}_{,2}= -3 - Δt = 2 - 0 = 2

Δx_{2} - Δx_{1} = [(0.5 + 0)/2 × 2] - [(-1.75 + (-3))/2 ×2] = 5.25

Two cars travel on the parallel lanes of a two-lane road. The cars are at the same location at time t=0s, and move in such a way as to produce the velocity (relative to the ground) vs. time graph shown in the figure. (Figure 1)On the graph, one vertical block is equivalent to one velocity unit.

Rank the distance between the cars at the lettered times.

Rank from largest to smallest. To rank items as equivalent, overlap them.

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Position & Velocity Graphs concept. If you need more Position & Velocity Graphs practice, you can also practice Position & Velocity Graphs practice problems.