Motion in two dimensions Name Mech HW-21 Some students used two straight segments and two half circles to make a track for a toy car, shown in the scale diagram at right. Each box on the diagram is 5 cm on a side. The car moves clockwise around the track at constant speed. It takes the car 8 s to go around the track once. a. Determine the speed of the car. Show your work. B 1 box = 5 cm b. Consider the interval that begins when the car passes point A and ends when the car passes point B (halfway around the track). i. On the diagram at right, draw an arrow to represent the direction of Direction of the average velocity over this interval. Explain your reasoning. from A to B avg ya ii. Determine the magnitude of the average velocity of the car over this interval. Show your work. Direction of a iii. On the diagram at right, draw an arrow to represent the direction of the average acceleration over this interval. Explain your reasoning. from A to B iv. Determine the magnitude of the average acceleration of the car over this interval. Show your work.
The correct answer and explanation is:
Solution and Explanation
(a) Determining the Speed of the Car
To find the speed of the car, we first determine the total distance it travels in one complete lap.
- Measure the Track Dimensions:
- The track consists of two straight segments and two half circles.
- Each box in the diagram represents 5 cm.
- From the diagram, each straight segment is 20 cm long.
- The two half-circles together form a full circle with a diameter of 10 cm, giving a circumference of: C=πd=π(10 cm)=31.4 cmC = \pi d = \pi (10 \text{ cm}) = 31.4 \text{ cm}
- Total Distance:
- The straight sections together contribute 40 cm.
- The circular sections contribute 31.4 cm.
- So, the total track length is: 40+31.4=71.4 cm40 + 31.4 = 71.4 \text{ cm}
- Find the Speed:
- Given that the car takes 8 seconds to complete one lap: v=Total DistanceTime=71.48=8.93 cm/sv = \frac{\text{Total Distance}}{\text{Time}} = \frac{71.4}{8} = 8.93 \text{ cm/s}
(b) Motion from A to B
Since the car moves at constant speed but changes direction, we analyze velocity and acceleration.
(i) Direction of the Average Velocity
- The car moves halfway around the track, from A to B.
- The average velocity is the displacement divided by time.
- Since A and B are opposite points on the track, the displacement vector points directly across the track, forming a straight line.
- On the diagram, the arrow should point directly from A to B, cutting through the track’s center.
(ii) Magnitude of the Average Velocity
- The displacement is simply the diameter of the track, which is 10 cm.
- The time to travel from A to B is half the total time: t=82=4 st = \frac{8}{2} = 4 \text{ s}
- The average velocity magnitude is: vavg=DisplacementTime=104=2.5 cm/sv_{\text{avg}} = \frac{\text{Displacement}}{\text{Time}} = \frac{10}{4} = 2.5 \text{ cm/s}
(iii) Direction of the Average Acceleration
- Acceleration depends on velocity change.
- The velocity at A and B are perpendicular to each other.
- The change in velocity points toward the track’s center.
- The average acceleration arrow should be drawn from the midpoint of A to B, pointing toward the center of the track.
(iv) Magnitude of the Average Acceleration
- Using vector change:
- Initial velocity: v⃗A\vec{v}_A (horizontal)
- Final velocity: v⃗B\vec{v}_B (vertical)
- Since both have the same magnitude, the change in velocity forms a right triangle.
- Magnitude of velocity change: ∣Δv∣=v2+v2=(8.93)2+(8.93)2=12.63 cm/s|\Delta v| = \sqrt{v^2 + v^2} = \sqrt{(8.93)^2 + (8.93)^2} = 12.63 \text{ cm/s}
- Average acceleration: aavg=∣Δv∣t=12.634=3.16 cm/s2a_{\text{avg}} = \frac{|\Delta v|}{t} = \frac{12.63}{4} = 3.16 \text{ cm/s}^2
Thus, the average acceleration has a magnitude of 3.16 cm/s² and points toward the center of the track.
Here’s the scale diagram of the toy car track, including the velocity and acceleration directions. Let me know if you need any modifications!
