Question

2.explain whether the size of an objects displacement could be greater than the distance the object travels.
3.describe the motion represented by a horizontal line on the distance time graph below.
4.describe the difference between average speed and constant speed.

1. during a car trip, can a cars instantaneous speed ever be greater than its average speed?

6.miko walked a distance of 1.60 km in 30 min. find her average speed in m/s.

1. a car travels at a constant speed of 30.0 m/s for 0.8 hr. find the total distance traveled in km.

Explanation:

Step1: Define displacement and distance

Displacement is the straight - line distance from the initial to the final position. Distance is the total path length traveled.

Step2: Analyze their relationship

Since displacement is the shortest path between two points and distance is the total path covered, the size of an object's displacement cannot be greater than the distance the object travels. It can be equal (when the object moves in a straight - line without changing direction) or less.

Step3: Analyze distance - time graph

On a distance - time graph, the slope represents speed. A horizontal line means the distance does not change with time. So the object is at rest.

Step4: Define average and constant speed

Average speed is total distance divided by total time ($v_{avg}=\frac{d_{total}}{t_{total}}$). Constant speed means the speed of the object remains the same throughout the motion.

Step5: Consider instantaneous and average speed

Instantaneous speed is the speed at a particular instant. During a car trip, if the car speeds up and slows down, the instantaneous speed can be greater than the average speed when the car is moving faster than its average rate.

Step6: Calculate Miko's average speed

First, convert distance to meters ($d = 1.60\ km=1600\ m$) and time to seconds ($t = 30\ min = 30\times60\ s=1800\ s$). Then use the formula $v=\frac{d}{t}$. So $v=\frac{1600}{1800}\approx0.89\ m/s$.

Step7: Calculate car's distance

First, convert time to seconds ($t = 0.8\ hr=0.8\times3600\ s = 2880\ s$). Use the formula $d = vt$, where $v = 30.0\ m/s$. So $d=30\times2880 = 86400\ m$. Convert to kilometers: $d = 86.4\ km$.

Answer:

1. No, the size of an object's displacement cannot be greater than the distance the object travels.
2. The object is at rest.
3. Average speed is total distance divided by total time. Constant speed means the speed remains the same throughout the motion.
4. Yes, during a car trip, a car's instantaneous speed can be greater than its average speed.
5. $0.89\ m/s$
6. $86.4\ km$