a body at rest is given an innital uniform acceleration of 8.0m/s-2 for 30s,acceleration reduce to 5.0m/s-2 for d next twenty sec.body mentains speed ten for 60s after brought to rest in 20s draw in graph 1.using d graph to calculate 2.maximum speed attain in motion 3.total distance travel during d first 50s 4.average speed during thesame interval as in (ii)
what is the question you are asking?
1) obviously can't
2) max is at end of 20 second acc
v = vo + at for 1st and second interval
3)x = 1/2(v + vo)t
again for two interval
4) total dist/total time
To solve the given problems, we need to analyze the given information step by step.
1. Using the graph to calculate:
To draw the graph, we need to first understand the motion of the body. From the given information:
a) The body experiences an initial uniform acceleration of 8.0 m/s^2 for 30 seconds.
b) The acceleration is then reduced to 5.0 m/s^2 for the next 20 seconds.
c) Afterward, the body maintains a constant speed for 60 seconds.
d) Finally, the body is brought to rest in 20 seconds.
Now, let's analyze the graphs for different periods of motion:
a) Initial acceleration period (0-30 seconds):
During this period, the body experiences a constant acceleration of 8.0 m/s^2. As a result, the velocity increases linearly with time, and the distance traveled increases quadratically.
The velocity-time graph (V vs. t) during this interval would be a straight line with a slope of 8.0 m/s^2. The position-time graph (s vs. t) would be a parabolic curve.
b) Reduced acceleration period (30-50 seconds):
During this period, the body experiences a constant acceleration of 5.0 m/s^2. Similar to the previous case, the velocity-time graph (V vs. t) would be a straight line with a slope of 5.0 m/s^2. The position-time graph (s vs. t) would also be a parabolic curve but with a different slope.
c) Constant speed period (50-110 seconds):
During this period, the body maintains a constant speed, which means it has zero acceleration. So, the velocity-time graph (V vs. t) would be a horizontal line. The position-time graph (s vs. t) would also be a straight line parallel to the time axis.
d) Deceleration period (110-130 seconds):
During this period, the body is brought to rest in 20 seconds, which means it experiences uniform deceleration. The velocity-time graph (V vs. t) would be a straight line with a negative slope, decreasing from the previous constant speed value to zero. The position-time graph (s vs. t) would be a parabolic curve with a negative slope.
Based on these descriptions, you can sketch the position-time (s vs. t) graph and the velocity-time (V vs. t) graph and use them to answer the given questions.
2. Maximum speed attained in motion:
To find the maximum speed, we need to look at the velocity-time (V vs. t) graph. The highest point on the graph represents the maximum speed attained by the body. You can determine this point from the vertex of the parabolic curve during the initial acceleration period (0-30 seconds).
3. Total distance traveled during the first 50 seconds:
To find the total distance traveled, we need to calculate the area under the position-time (s vs. t) graph from 0 to 50 seconds. Divide the graph into different regions (acceleration, reduced acceleration, constant speed, and deceleration) and calculate the area separately. Then, add these areas to get the total distance traveled during the first 50 seconds.
4. Average speed during the same interval as in (ii):
To find the average speed, divide the total distance traveled during the first 50 seconds by the time taken (50 seconds).
By following these steps, you will be able to calculate the required values and solve the given problems using the provided graph.