AP Physics 1: 1.1 Position, Velocity, and Acceleration – Exam Style questions with Answer- FRQ

Ans:

a.

\( \begin{gathered}v=\frac{dx}{d t} \\ \int v_{d t}=\int d x \\ \int v_{d t}= Area~ under~ (v-t) ~graph
\\ \Rightarrow \quad 2 \times 4+\frac{1}{2} 0.5 \times 4+\frac{1}{2} \frac{1}{2}(-4) \\ x_f-x_i=8+1-1\\x_f=8+3.5=11.5 m\end{gathered} \)

b.

student is moving in one direction up to 2.5 second,

So maximum positive position will be $x_f-3.5=2 \times 4+\frac{1}{2} 0.5 \times 4\Rightarrow x_f=12.5 m $

c.

\( \begin{aligned}\langle v\rangle & =\frac{\Delta x}{\Delta t} \\ \Delta x & =\text { Area } \\ & =\frac{8+1-1}{3} \\ & =2.67 \mathrm{~m} / \mathrm{s}\end{aligned} \)

d.

average speed is total speed (modulus of displacement) / total time taken 

if we draw speed time graph and then calculate the area we will get the distance covered

\(\begin{aligned}
\langle\bar{v}\rangle & =\frac{8+1+1}{3} \\
& =3.33 \mathrm{m/s}
\end{aligned}\)

e .

\( \begin{aligned}\langle a\rangle & =\frac{v_f-v_i}{t_f-t_i} \\ & =\frac{0-4}{2.5-2} \\ & =-8 \mathrm{~m} / \mathrm{s}^2\end{aligned} \)

f.

\( \begin{aligned}\langle a\rangle & =\frac{v_f-v_i}{t_f-t_i} \\ & =\frac{-4-0}{3-2.5} \\ & =-8 \mathrm{~m} / \mathrm{s}^2\end{aligned} \)