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🎡AP Physics 1
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🎡AP Physics 1

FRQ 2 – Translation Between Representations
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Unit 1: Kinematics
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Practice FRQ 1 of 191/19

2. A student rolls a cart along a straight, level track that lies on the floor in a laboratory.

Figure 1. Velocity v_x,A versus time t for the cart, measured in the laboratory frame (observer A).

Figure 1

Figure 2. Blank axes for sketching position x_A versus time t (observer A), from t = 0 to t = 6.0 s.

Figure 2
A.

Draw a single continuous curve on Figure 2 that represents the cart’s position xAx_AxA​ as a function of time from t=0t=0t=0 to t=6.0 st=6.0\ \text{s}t=6.0 s as measured by observer A. Figure 1 shows the cart’s velocity vx,Av_{x,A}vx,A​ as a function of time as measured by observer A from t=0t=0t=0 to t=6.0 st=6.0\ \text{s}t=6.0 s. Figure 2 provides blank axes for a position graph.

• The cart starts at xA=0x_A=0xA​=0 at t=0t=0t=0.
• Your curve must be consistent with the velocity–time graph in Figure 1.
• Clearly indicate changes in concavity where appropriate.

Figure 3. Two inertial observers and the +x direction: observer A in the lab and observer B on a platform moving at constant speed +0.80 m/s relative to the lab.

Figure 3
B.

Starting with the definition of relative velocity in one dimension, derive an expression for the cart’s average velocity as measured by observer B, v‾x,B\overline{v}_{x,B}vx,B​, over the time interval from t=0t=0t=0 to t=6.0 st=6.0\ \text{s}t=6.0 s. Express your final answer in terms of v‾x,A\overline{v}_{x,A}vx,A​ (the cart’s average velocity measured by observer A over the same interval) and vB/A=+0.80 m/sv_{B/A}=+0.80\ \text{m/s}vB/A​=+0.80 m/s. Begin your derivation by writing a fundamental physics principle or an equation from the reference information. Observer B moves at constant velocity +0.80 m/s+0.80\ \text{m/s}+0.80 m/s along the xxx-axis relative to observer A, as shown in Figure 3. At t=0t=0t=0 the observers’ origins coincide.

Figure 4. Blank axes for sketching the projectile trajectory y(x) after the cart leaves the track (origin at the track edge).

Figure 4
C.

At t=6.0 st=6.0\ \text{s}t=6.0 s the cart leaves the end of the track at height 1.20 m1.20\ \text{m}1.20 m above the floor and then moves as a projectile. The cart’s horizontal velocity at the instant it leaves the track is equal to the value of vx,Av_{x,A}vx,A​ at t=6.0 st=6.0\ \text{s}t=6.0 s from Figure 1. Take g=9.8 m/s2g=9.8\ \text{m/s}^2g=9.8 m/s2.

i.

Sketch and label a curve on Figure 4 that represents the trajectory of the cart during projectile motion, y(x)y(x)y(x), from the instant it leaves the track until it reaches the floor.

ii.

Sketch and label on a separate diagram the perpendicular components of the cart's instantaneous velocity vector at the instant it leaves the track: vxv_xvx​ (horizontal component) and vyv_yvy​ (vertical component). Clearly indicate both magnitude and direction using arrows.

D.

Indicate whether the magnitude of the cart’s horizontal displacement during the time it is in the air, as measured by observer B, ∣ΔxB∣|\Delta x_B|∣ΔxB​∣, is greater than, less than, or equal to the magnitude of the cart’s horizontal displacement during the time it is in the air, as measured by observer A, ∣ΔxA∣|\Delta x_A|∣ΔxA​∣. Use the information from Figure 1 and the following values: the cart leaves the track at t=6.0 st=6.0\ \text{s}t=6.0 s from a height of 1.20 m1.20\ \text{m}1.20 m, observer B moves at vB/A=+0.80 m/sv_{B/A}=+0.80\ \text{m/s}vB/A​=+0.80 m/s, and g=9.8 m/s2g=9.8\ \text{m/s}^2g=9.8 m/s2. Assume the cart’s vertical velocity is 0 m/s0\ \text{m/s}0 m/s at the instant it leaves the track.

∣ΔxB∣>∣ΔxA∣|\Delta x_B| > |\Delta x_A|∣ΔxB​∣>∣ΔxA​∣
∣ΔxB∣<∣ΔxA∣|\Delta x_B| < |\Delta x_A|∣ΔxB​∣<∣ΔxA​∣
∣ΔxB∣=∣ΔxA∣|\Delta x_B| = |\Delta x_A|∣ΔxB​∣=∣ΔxA​∣
Justify how your response is consistent with the velocity components you sketched in part C and the idea of different inertial reference frames.

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Free Response Question Practice

This practice environment simulates the AP AP Physics 1 Free Response Questions section. Here are some guidelines:

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