Question

practice: motion graphs - gentry
part 1: match the letters on the graph to the statements in #1 - 3 that best describe claras motion. then answer questions 4 and 5.

1. clara stops for 10 minutes to catch up with a friend.
2. clara jogs 600 m in 5 minutes.
3. she walks at a constant speed of 80 m/min.
4. is clara jogging faster during part a of her run or part c? refer to the model and justify your answer with reasoning.
5. calculate claras average speed for her entire journey.

part 2: use the graph below to answer questions #6 - 10 about jays motion on his walk.

1. find jays average speed for his entire walk.
2. what is happening at part b?
3. find jays average speed for just part d.
4. compare jays motion at part a vs part d.
5. challenge: what do you think part c might represent?

Explanation:

Step1: Recall average - speed formula

The formula for average speed is $v = \frac{d}{t}$, where $v$ is average speed, $d$ is total distance, and $t$ is total time.

Step2: Analyze Clara's motion

• From Clara's graph, the total distance $d_{Clara}=1400$ m and the total time $t_{Clara}=25$ min.
• Using the average - speed formula $v_{Clara}=\frac{d_{Clara}}{t_{Clara}}=\frac{1400}{25}=56$ m/min.

Step3: Analyze Jay's motion

• From Jay's graph, the total distance $d_{Jay}=10$ m and the total time $t_{Jay}=45$ s.
• Using the average - speed formula $v_{Jay}=\frac{d_{Jay}}{t_{Jay}}=\frac{10}{45}=\frac{2}{9}\approx0.22$ m/s.

Step4: Analyze Jay's part D

• In part D of Jay's motion, the distance covered $d_D = 10 - 3=7$ m and the time taken $t_D=45 - 30 = 15$ s.
• Using the average - speed formula $v_D=\frac{d_D}{t_D}=\frac{7}{15}\approx0.47$ m/s.

Step5: Compare Jay's part A and D

• In part A of Jay's motion, the distance covered $d_A = 6$ m and the time taken $t_A=15$ s, so $v_A=\frac{d_A}{t_A}=\frac{6}{15}=0.4$ m/s.
• In part D, $v_D\approx0.47$ m/s. So Jay moves faster in part D than in part A.

Step6: Interpret part C of Jay's motion

Part C might represent Jay walking back towards the starting point as the distance from the starting - point is decreasing.

Answer:

1. Clara is jogging faster in part C. In part A, if we assume the starting point is $(0,0)$ and an end - point on part A say $(5,400)$ (from the graph), the speed $v_A=\frac{400}{5}=80$ m/min. In part C, the distance covered is $1400 - 900 = 500$ m in $20 - 15=5$ min, so the speed $v_C=\frac{500}{5}=100$ m/min.
2. 56 m/min
3. $\frac{2}{9}\approx0.22$ m/s
4. Jay has stopped.
5. $\frac{7}{15}\approx0.47$ m/s
6. Jay moves faster in part D. In part A, speed $v_A = 0.4$ m/s and in part D, speed $v_D\approx0.47$ m/s.
7. Jay walking back towards the starting point.