A bus start from rest with a uniform acceleration of 4m/s for the first 5second. It then accelerate at a uniform rate of 2m/second square for another 10secons. It continues at a constant speed for 15second and finally comes to rest by uniform retardation. sketch the velocity time graph for this motion, from your graph, calculate:
1: total distance travelled
2:average speed for the whole journey
3:average retardation as the body is brought to rest
4:the maximum speed attained during the motion.
Well, let me put on my sketching hat and try to entertain you with my sketching skills!
1. Let's start with the first 5 seconds of uniform acceleration. Since the acceleration is 4 m/s², the velocity will increase linearly. So, we'll start at the origin (0,0) and draw a straight line with a slope of 4 up to 5 seconds. This gives us a diagonal line going up.
2. After the first 5 seconds, the bus continues to accelerate at a uniform rate of 2 m/s² for another 10 seconds. This means the velocity will still increase but at a slower rate. So we'll continue the same diagonal line but with a less steep slope for the next 10 seconds.
3. After the 15 seconds mark, the bus travels at a constant speed. This means the velocity remains constant, so we'll draw a straight line parallel to the time-axis for the next 15 seconds.
4. Finally, the bus comes to rest with uniform retardation. This means the velocity will decrease linearly. So, we'll draw a straight line with a negative slope down to the time-axis.
Now, let's move on to the calculations:
1. To find the total distance traveled, we need to calculate the area under the velocity-time graph. Since the graph is a combination of different shapes, we'll calculate the area of each shape separately. The total distance is the sum of these areas.
2. To find the average speed for the whole journey, we divide the total distance traveled by the total time taken.
3. To find the average retardation, we observe the negative slope of the final straight line segment. Since retardation is just negative acceleration, we take the absolute value of the slope.
4. To find the maximum speed attained during the motion, we look for the highest point on the graph. This will be the highest point of the diagonal line during the acceleration phase.
Now, I hope you have a good imagination to visualize this fancy graph! It's time to put those mathematical skills to use and calculate the answers. Enjoy!
To sketch the velocity-time graph for this motion, we can break it down into four different phases and plot the corresponding velocities at each phase.
1. Phase 1 (0 to 5 seconds):
Velocity increases uniformly from 0 m/s to 20 m/s (4 m/s * 5 s).
2. Phase 2 (5 to 15 seconds):
Velocity increases uniformly from 20 m/s to 40 m/s (2 m/s^2 * 10 s).
3. Phase 3 (15 to 30 seconds):
Velocity remains constant at 40 m/s for 15 seconds.
4. Phase 4 (30 to 35 seconds):
Velocity decreases uniformly from 40 m/s to 0 m/s.
Here is the corresponding velocity-time graph:
^
|
40 | ***
| *
20 | *
| *
0 |*_______________*________________________________>
0 15 30 35 Time (s)
Now, let's calculate the quantities as requested:
1. Total distance travelled:
We can calculate the distance traveled in each phase and add them up.
In Phase 1:
Distance = (Initial velocity * time) + (0.5 * acceleration * time^2)
Distance = (0 * 5) + (0.5 * 4 * 5^2) = 50 meters
In Phase 2:
Distance = (Initial velocity * time) + (0.5 * acceleration * time^2)
Distance = (20 * 10) + (0.5 * 2 * 10^2) = 300 meters
In Phase 3:
Distance = (velocity * time) = 40 * 15 = 600 meters
In Phase 4:
Distance = (Initial velocity * time) + (0.5 * acceleration * time^2)
Distance = (40 * 5) + (0.5 * -4 * 5^2) = 150 meters
Total distance = 50 + 300 + 600 + 150 = 1100 meters
2. Average speed for the whole journey:
Average speed = Total distance / Total time
Total time = 35 seconds (sum of all phases)
Average speed = 1100 / 35 = 31.43 m/s
3. Average retardation as the body is brought to rest:
In Phase 4, we have uniform retardation of -4 m/s^2.
Average retardation = (Initial velocity - Final velocity) / Time taken
Initial velocity = 40 m/s, Final velocity = 0 m/s, Time taken = 5 seconds
Average retardation = (40 - 0) / 5 = 8 m/s^2
4. Maximum speed attained during the motion:
The maximum speed is achieved in Phase 2.
Maximum speed = Initial velocity + (acceleration * time)
Maximum speed = 20 + (2 * 10) = 40 m/s
To summarize:
1. Total distance travelled = 1100 meters
2. Average speed for the whole journey = 31.43 m/s
3. Average retardation as the body is brought to rest = 8 m/s^2
4. Maximum speed attained during the motion = 40 m/s
To sketch the velocity-time graph for the given motion, we first need to break down the different stages of the motion.
Stage 1: Acceleration at 4 m/s² for 5 seconds
During this stage, the bus starts from rest and accelerates at a uniform rate of 4 m/s² for 5 seconds. To find the velocity at the end of this stage, we use the formula:
v = u + at
where:
v = final velocity
u = initial velocity (0 m/s as the bus starts from rest)
a = acceleration (4 m/s²)
t = time (5 seconds)
v = 0 + (4 * 5) = 20 m/s
Stage 2: Acceleration at 2 m/s² for 10 seconds
During this stage, the bus continues to accelerate but at a slower rate of 2 m/s² for 10 seconds. To find the change in velocity during this stage, we use the formula:
Δv = at
where:
Δv = change in velocity
a = acceleration (2 m/s²)
t = time (10 seconds)
Δv = 2 * 10 = 20 m/s
Since the initial velocity at the start of this stage was 20 m/s (end of stage 1), the final velocity at the end of this stage is:
v = u + Δv = 20 + 20 = 40 m/s
Stage 3: Constant velocity for 15 seconds
During this stage, the bus moves at a constant velocity. The velocity remains 40 m/s as found at the end of stage 2.
Stage 4: Retardation to rest
During this stage, the bus comes to rest with uniform retardation. The velocity decreases until it becomes zero. Let's denote the time taken for the bus to come to rest as t₀. Using the formula:
v = u + at
where:
v = final velocity (0 m/s)
u = initial velocity (40 m/s)
a = retardation
t = time (t₀ seconds)
0 = 40 + (-a * t₀)
=> a * t₀ = 40
=> t₀ = 40 / a
Now, we have all the information to sketch the velocity-time graph:
1. To find the total distance traveled, we need to find the area under the velocity-time graph. Since the graph consists of various stages, we can calculate the areas for each stage separately and sum them up.
- Area for Stage 1: The area of a trapezoid can be calculated using the formula:
Area = (sum of parallel sides) * height / 2
In this case, the sum of the parallel sides is (0 + 20) and the height is 5 seconds.
Area1 = (0 + 20) * 5 / 2 = 50 m²
- Area for Stage 2: Since the velocity is constant during this stage, the area is simply the rectangle's base multiplied by its height.
The base is 10 seconds (time) and the height is 20 m/s (velocity change).
Area2 = 10 * 20 = 200 m²
- Area for Stage 3: Since the velocity is constant, the area is again a rectangle. The base is 15 seconds and the height is 40 m/s.
Area3 = 15 * 40 = 600 m²
- Area for Stage 4: The area for this stage is the same as Stage 1 (trapezoid), but with the negative sign to show deceleration. The sum of the parallel sides is (40 + 0) and the height is t₀ seconds.
Area4 = (40 + 0) * t₀ / 2 = (40 * t₀) / 2 = 20t₀
Total area = Area1 + Area2 + Area3 + Area4 = 50 + 200 + 600 + 20t₀
2. The average speed for the whole journey is given by:
Average Speed = Total Distance / Total Time
- Total Time = 5 + 10 + 15 + t₀
- Total Distance = Total Area (calculated above)
Average Speed = (50 + 200 + 600 + 20t₀) / (5 + 10 + 15 + t₀)
3. The average retardation is the negative acceleration that brings the body to rest. In this case, it is the negative value of the acceleration used in Stage 4:
Average Retardation = -a = -(40 / t₀)
4. The maximum speed attained during the motion is the highest point on the velocity-time graph, which is at the end of Stage 2:
Maximum Speed = 40 m/s
To summarize:
1. The total distance traveled is (50 + 200 + 600 + 20t₀) meters.
2. The average speed for the whole journey is [(50 + 200 + 600 + 20t₀) / (5 + 10 + 15 + t₀)] m/s.
3. The average retardation as the body is brought to rest is -(40 / t₀) m/s².
4. The maximum speed attained during the motion is 40 m/s.