a body at rest is given an initial uniform acceleration of 8.0m/s for 30. after which the acceleration is reduced to 5.0s for the next 20s the body maintains the speed attained for 60s.After which it is brought to rest in 20s.Draw the velocity-city graph of the motion using the information given above? using the graph calculate,the maximum speed attained during the motion,the total distance traveled during the first 50s and the average speed during the same interval in 3?
The velocity-time graph is shown below:
Velocity (m/s) | 0 | 8 | 16 | 24 | 30 | 35 | 40 | 45 | 50 | 45 | 40 | 35 | 30 | 25 | 20 | 15 | 10 | 5 | 0
Time (s) | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 | 110 | 120 | 130 | 140 | 150 | 160 | 170 | 180
Maximum speed attained during the motion: 45 m/s
Total distance traveled during the first 50s: 400 m
Average speed during the same interval: 8 m/s
a body at rest is given an initial uniform acceleration of 8.0m/s for 30. after which the acceleration is reduced to 5.0s for the next 20s the body maintains the speed attained for 60s.After which it is brought to rest in 20s.Draw the velocity-city graph of the motion using the information given above? using the graph calculate,the maximum speed attained during the motion,the total distance traveled during the first 50s and the average speed during the same interval in 3?
a body at rest is given an initial uniform acceleration of 8.0m/s for 30. after which the acceleration is reduced to 5.0s for the next 20s the body maintains the speed attained for 60s.After which it is brought to rest in 20s.Draw the velocity-city graph of the motion using the information given above? using the graph calculate,the maximum speed attained during the motion,the total distance traveled during the first 50s and the average speed during the same interval in 3? With working
To draw the velocity-time graph for the given motion, we will divide the motion into different intervals and calculate the velocity for each interval.
Interval 1: Initial Acceleration (0 - 30 seconds)
During this interval, the body undergoes uniform acceleration of 8.0 m/s² for 30 seconds. We can use the equation of motion:
v = u + at
where:
v = final velocity
u = initial velocity (which is 0 m/s as the body is at rest)
a = acceleration
t = time
Substituting the values:
v = 0 + 8.0(30)
v = 240 m/s
Interval 2: Reduced Acceleration (30 - 50 seconds)
In this interval, the acceleration is reduced to 5.0 m/s² for 20 seconds. We can again use the same equation of motion:
v = u + at
Substituting the values:
v = 240 + 5.0(20)
v = 340 m/s
Interval 3: Constant Velocity (50 - 110 seconds)
During this interval, the body maintains a constant speed. We're given that the speed attained in the previous interval (i.e., Interval 2) is maintained for 60 seconds, so the velocity remains constant at 340 m/s.
Interval 4: Deceleration (110 - 130 seconds)
In this interval, the body is brought to rest in 20 seconds. The final velocity is 0 m/s, and the initial velocity is 340 m/s (from Interval 3). Again, using the equation of motion:
v = u + at
Substituting the values:
0 = 340 + (-a)(20)
a = 170/20
a = -8.5 m/s² (negative because it's deceleration)
Using this information, we can now draw the velocity-time graph:
| /
340 -| /
| /
v -|/
+------------------
50 110 130
Now, let's use the graph to calculate the maximum speed attained, total distance traveled in the first 50 seconds, and the average speed during the same interval.
1. Maximum speed attained during the motion:
From the graph, we can see that the maximum speed attained is 340 m/s.
2. Total distance traveled during the first 50 seconds:
To find the total distance, we need to calculate the area under the velocity-time graph for the interval 0-50 seconds. Since the graph is a rectangle, the area is equal to the product of the base (time) and height (velocity).
Area = base * height
= 50 * 340
= 17,000 m
So, the total distance traveled during the first 50 seconds is 17,000 meters.
3. Average speed during the same interval:
Average speed is given by the total distance traveled divided by the total time taken. In this case, the total time is 50 seconds.
Average speed = Total distance / Total time
= 17,000 m / 50 s
= 340 m/s
Therefore, the average speed during the first 50 seconds is 340 m/s.