A body at rest is given an initial uniform acceleration of 8m/s2 for 30s after which the acceleration is reduced to 5m/s2 for the next twenty seconds. The body maintains the speed attained for 60s, after which it is brought to rest in 20s.
1. Draw the velocity time graph of the motion using the information given.
2. Using the graph, calculate the maximum speed attained during the motion
3. Average retardation as the body is being brought to rest
4. Total distance travelled during the first 50s
5. Average speed during the first 50s
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1)v1=u+a1t1=0+(8*3)=240m/s
2)v2=v1+a2t2=240+(-5)*20=140m/s
3)total area convered in 50s=Area of motion in 30s+Area of motion in 20s=area of triangle + area of parallelogram=1/2*30*240+1/2(240+140)*20=7400m
4)average speed=total distance covered in 50s/50s=7400/50=148m/s
I dont know help me please
1.u=0
V=u+at=0+8*30=240
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To answer these questions, we will first need to plot the velocity-time graph of the motion using the information given. Once we have the graph, we can use it to calculate the maximum speed attained, average retardation, total distance traveled during the first 50s, and the average speed during the first 50s.
1. Plotting the velocity-time graph:
To plot the graph, we will divide the given time intervals and calculate the corresponding velocities for each interval.
Start with the initial velocity (u) as zero since the body is at rest initially.
- From t = 0 to t = 30s, the body undergoes uniform acceleration of 8 m/s^2, so the slope of the graph will be a straight line with a positive slope of 8 m/s^2 for 30 seconds. The final velocity (v) can be calculated using the equation v = u + at, where u = 0, a = 8 m/s^2, and t = 30s.
- From t = 30s to t = 50s, the acceleration reduces to 5 m/s^2. The slope of the graph will be a straight line with a positive slope of 5 m/s^2 for 20 seconds. The final velocity (v) can again be calculated using the equation v = u + at, where u is the velocity at the end of the previous interval, which is 8 m/s^2, a = 5 m/s^2, and t = 20s.
- From t = 50s to t = 110s, the body maintains a constant speed, so the graph will be a straight line parallel to the time axis. The speed can be obtained from the previous interval, which is 5 m/s.
- Finally, it takes 20s to bring the body to rest, so the graph will be a straight line with a negative slope of -5 m/s^2 for 20 seconds, intersecting the time axis at t = 130s.
2. The maximum speed attained during the motion:
To find the maximum speed, we need to identify the highest point on the velocity-time graph. In this case, the highest point is where the graph levels off and becomes parallel to the time axis. This occurs at t = 50s, where the speed is 5 m/s.
3. Average retardation as the body is being brought to rest:
To calculate the average retardation, we need to determine the rate at which the velocity decreases during the last 20s. From t = 110s to t = 130s, the velocity changes from 5 m/s to 0 m/s. The decrease in velocity is 5 m/s, and the time taken is 20s. Therefore, the average retardation is given by the equation a = (change in velocity)/(time taken) = 5 m/s divided by 20s.
4. Total distance traveled during the first 50s:
To calculate the total distance traveled during the first 50s, we need to find the area under the velocity-time graph for this time interval. Since the graph consists of a straight line with a positive slope, the area will be a triangle. The formula for the area of a triangle is A = (base * height) / 2, where the base is the time interval (50s) and the height is the velocity (5 m/s).
5. Average speed during the first 50s:
To find the average speed, we need to divide the total distance traveled during the first 50s by the time taken (50s).
By following these steps, we can answer the questions using the information given and the velocity-time graph of the motion.