Displacement, Velocity, and Acceleration AP Physics 1 FRQ

Displacement, Velocity, and Acceleration AP Physics 1 FRQ – Exam Style Questions etc.

Displacement, Velocity, and Acceleration AP Physics 1 FRQ

Unit: 1. Kinematics

Weightage : 10-15%

AP Physics 1 Exam Style Questions – All Topics

Question

The horizontal velocity of a student moving in a hallway is shown in the graph above. At $t=0$, the student is at $x=+3.5 \mathrm{~m}$.
(a) Determine the position of the student at $t=3 \mathrm{~s}$.
(b) What is the most positive position the student reaches?
(c) Determine the average velocity of the student between $t=0$ and $t=3 \mathrm{~s}$ .
(d) Determine the average speed of the student between $t=0$ and $t=3 \mathrm{~s}$
(e) Determine the average acceleration of the student between $2 \mathrm{~s}$ and $2.5 \mathrm{~s}$
(f) Determine the average acceleration of the student between $2.5 \mathrm{~s}$ and $3.0 \mathrm{~s}$

▶️Answer/Explanation

Ans:

$ \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} $

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 $

$ \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} $

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} $

$ \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} $

Question

In the 2016 Olympics, American Ryan Crouser won gold in the shot put. He won with an Olympic record distance of $22.52 \mathrm{~m}$. The shot put was released at an angle of $38.2^{\circ}$ with a velocity of $14.3 \mathrm{~m} / \mathrm{s}$.

a. What height above the ground was the shot put released?

b.) i. How fast was the shot put moving when it hit the ground?

ii. What angle did the shot put hit the ground?

c. What maximum height did the shot put reach above the ground while it was in the air?

d. What is the speed of the shot put when it reaches the highest point?

▶️Answer/Explanation

Ans:

$ \begin{gathered}x=\left(u_0\right)_x t+\frac{1}{2} a_x t^2 \\ 22.52=11.2 t+0 \\ t=2.0~ \mathrm{sec}\end{gathered} $

$ \begin{aligned} y_{-} y_0 & =\left(u_0\right)_y t+\frac{1}{5} a_y t^2 \\ 0-y_0 & =8.24(2)+\frac{1}{5}(-9.8)(2)^2 \\ y_0 & =1.92 \mathrm{~m}\end{aligned} $

b(i).

$\begin{aligned} & V_y^2=V_{y_0}{ }^2+2 a_y\left(y-y_0\right) \\ & V_y^2=(8.84)^2+2(-9.8)(-1.92) \\ & \sqrt{V_y^2}=\sqrt{115.7776} \\ & V_y=10.76 \mathrm{~m} / \mathrm{s}\end{aligned}$

b(ii).

$ \begin{aligned} \tan \theta & =\frac{10.76}{11.2} \\ \theta & =\tan ^{-1}(0.96) \\ \theta & =43.8^{\circ}\end{aligned} $

Maximum height will be at half of time of flight of its motion

$\begin{aligned} & y=y_0+v_{y_0} t+\frac{1}{2} a_x t^2 \\ & y=1.92+8.84 t+\frac{1}{2}(-9.8) t^2 \\ & y=-4.9 t^2+8.84 t+1.92 \\ & y=-4.9(1)^2+8.84(1)+1.92 \\ & y=5.86 \text { meters }\end{aligned}$

d. At max. height vertical component of velocity will be zero , only horizontal will be present so , speed $=11.2 m/s$

Displacement, Velocity, and Acceleration AP Physics 1 FRQ