A car starts at rest from one station A and is uniformly accelerated until it reaches a speed of 25 m/s. It then travels with this speed until the brakes are applied and brought the car to rest with a uniform retardation at the next station B. The magnitude of the retardation is three times the magnitude of the acceleration. The distance between the two station is 8 km and the time taken for the journey is 6 minutes.
(a) Draw V – t graph for the information
(b) From the graph find the distance travelled at the constant speed
(c) The time taken at the constant speed.
(d) the initial acceleration.
(a) a sloping line up to v=25, a horizontal segment, a segment sloping back down to v=0
(b) distance = speed * time
(c) the length of the horizontal segment
(d) the slope of the first section.
Solution
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To answer the questions, let's go step by step:
(a) Draw the V-t graph for the information:
The V-t (velocity - time) graph will have a few distinct sections.
1. The car starts from rest at station A and uniformly accelerates until it reaches a speed of 25 m/s. This can be represented by a straight line with positive slope, starting from the origin and reaching a point on the y-axis corresponding to 25 m/s.
2. After reaching this speed, the car travels with a constant speed of 25 m/s until station B. This section will be a straight horizontal line parallel to the x-axis.
3. At station B, the brakes are applied, and the car is brought to rest with a uniform retardation. This can be represented by a straight line with negative slope, starting from the point on the y-axis corresponding to 25 m/s and reaching the origin.
Thus, the V-t graph will have a "V" shape.
(b) From the graph, find the distance travelled at constant speed:
To find the distance travelled at constant speed, we need to find the area under the graph during the constant speed section. In this case, the area will be the rectangle formed by the horizontal line (constant speed) and the time axis (t).
Given that the distance between the two stations is 8 km and the time taken for the journey is 6 minutes, we need to convert the time to seconds and calculate the distance:
Distance = Speed × Time
The speed during the constant speed section is 25 m/s, and the time is 6 minutes, which is equal to 6 × 60 = 360 seconds.
Distance = 25 m/s × 360 s = 9000 meters = 9 km
So, the distance traveled at the constant speed is 9 km.
(c) Find the time taken at the constant speed:
The constant speed section on the V-t graph is a horizontal line. We can find the time taken by calculating the difference between the x-coordinates of the two points where this line intersects the graph.
In this case, the time taken at the constant speed is the difference between the two times when the velocity is 25 m/s. Since it's a horizontal line, the time difference will be the same as the time taken for the whole journey.
Given that the time taken for the journey is 6 minutes, or 6 × 60 = 360 seconds, the time taken at the constant speed is 360 seconds or 6 minutes.
(d) Find the initial acceleration:
We are given that the magnitude of the retardation is three times the magnitude of the acceleration.
Since the car starts from rest and accelerates until reaching 25 m/s, the acceleration can be calculated using the equation:
V^2 = U^2 + 2as
where V is the final velocity (25 m/s), U is the initial velocity (0 m/s), a is the acceleration, and s is the distance (the total distance between the two stations, which is 8 km or 8000 meters)
(25^2) = (0^2) + 2a(8000)
625 = 16000a
a = 625 / 16000
a = 0.039 m/s^2
Therefore, the initial acceleration is approximately 0.039 m/s^2.