Educational worksheet featuring velocity-time graphs and related questions for physics learning

Based on the "Velocity-Time Graphs Worksheet," I will solve the problems presented by analyzing the provided velocity-time graph. The graph illustrates the motion of a car over time, showing three distinct phases: acceleration, constant velocity, and deceleration.

Here is the step-by-step solution to the questions:

1. Acceleration between 0.4h and 0.6h

The question asks for the acceleration during the time interval from 0.4 hours to 0.6 hours.
- Looking at the graph, during this period, the velocity line is horizontal.
- A horizontal line on a velocity-time graph indicates that the velocity is constant (not changing).
- Acceleration is defined as the rate of change of velocity with respect to time ($a = \frac{\Delta v}{\Delta t}$).
- Since the change in velocity ($\Delta v$) is zero during this interval, the acceleration is also zero.

Answer: The acceleration between 0.4h and 0.6h is 0 km/h².

2. Deceleration between 0.6h and 1.0h

The question asks for the deceleration during the time interval from 0.6 hours to 1.0 hours.
- From the graph, we can see that the car's velocity decreases from 60 km/h to 0 km/h during this time.
- The change in velocity ($\Delta v$) is $0 - 60 = -60$ km/h.
- The change in time ($\Delta t$) is $1.0 - 0.6 = 0.4$ hours.
- Deceleration is the magnitude of negative acceleration, so we calculate the acceleration first:
$a = \frac{\Delta v}{\Delta t} = \frac{-60 \text{ km/h}}{0.4 \text{ h}} = -150 \text{ km/h}^2$
- The deceleration is the absolute value of this acceleration.

Answer: The deceleration between 0.6h and 1.0h is 150 km/h².

3. Total Distance Traveled

The question asks for the total distance traveled by the car.
- On a velocity-time graph, the distance traveled is equal to the area under the curve.
- The graph forms a trapezoid. We can calculate the area of this trapezoid using the formula:
$\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
- In this case, the two parallel sides are the top and bottom of the trapezoid. The bottom side (total time) is 1.0 hour, and the top side (time at constant velocity) is $0.6 - 0.4 = 0.2$ hours. The height is the maximum velocity, which is 60 km/h.
- So, the area (distance) is:
$\text{Distance} = \frac{1}{2} \times (1.0 + 0.2) \times 60 = \frac{1}{2} \times 1.2 \times 60 = 0.6 \times 60 = 36 \text{ km}$

Alternatively, we can break the area into three parts: a triangle (acceleration phase), a rectangle (constant velocity phase), and another triangle (deceleration phase).

- Acceleration Phase (0 to 0.4h):
Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.4 \times 60 = 12 \text{ km}$

- Constant Velocity Phase (0.4 to 0.6h):
Area of rectangle = $\text{length} \times \text{width} = 0.2 \times 60 = 12 \text{ km}$

- Deceleration Phase (0.6 to 1.0h):
Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.4 \times 60 = 12 \text{ km}$

- Total Distance:
$12 + 12 + 12 = 36 \text{ km}$

Answer: The total distance traveled is 36 km.