(a) A body at rest is given an initial uniform acceleration of 8.0 metre per second square for 30 seconds after which the acceleration is reduced to 5.0 metre per second square for the next 20 seconds. The body maintains the speed attained for 60 seconds after which it is brought to rest in 20 seconds.
Draw the velocity time graph of the motion using the information above.
(b) Using the graph calculate:
(i) Maximum speed attained during the motion
(ii) Average retardation as the body is being brought to rest.
(iii) Total distance travelled during the first 50 seconds.
(iv) Average speed during the same interval as in (iii).
To draw the velocity-time graph, we need to analyze the given information and plot the points accordingly.
(a) Here is the breakdown of the motion based on the given information:
1. Initial uniform acceleration of 8.0 m/s^2 for 30 seconds.
2. Acceleration reduced to 5.0 m/s^2 for the next 20 seconds.
3. Constant velocity maintained for 60 seconds.
4. Brought to rest in 20 seconds.
To draw the velocity-time graph, we divide the motion into different segments and analyze each segment separately.
Segment 1: Acceleration of 8.0 m/s^2 for 30 seconds:
In this segment, the velocity increases linearly with time. We can calculate the change in velocity (Δv) using the formula: Δv = a * t, where a is the acceleration and t is time. The initial velocity (u) is zero since the body is at rest.
Δv = 8.0 m/s^2 * 30 s = 240 m/s.
So, at the end of 30 seconds, the velocity becomes 240 m/s.
Segment 2: Acceleration reduced to 5.0 m/s^2 for the next 20 seconds:
Similar to segment 1, the velocity increases linearly with time, but with a reduced acceleration. Again, the initial velocity (u) is 240 m/s (as the velocity from the previous segment is carried forward).
Δv = 5.0 m/s^2 * 20 s = 100 m/s.
So, at the end of 50 seconds (30 + 20), the velocity becomes 240 m/s + 100 m/s = 340 m/s.
Segment 3: Constant velocity of 340 m/s for 60 seconds:
Here, the velocity remains constant, so there is no acceleration. The velocity remains 340 m/s for the entire 60 seconds.
Segment 4: Brought to rest in 20 seconds:
The velocity decreases linearly to zero with constant deceleration. The initial velocity (u) is 340 m/s (as the velocity from the previous segment is carried forward).
Using the formula v = u + at (where v = final velocity, u = initial velocity, a = acceleration, and t = time), we can find the deceleration (retardation) during this segment.
Final velocity (v) = 0 m/s, initial velocity (u) = 340 m/s, time (t) = 20 s.
0 = 340 m/s + a * 20 s.
Solving for a, we get a = -17 m/s^2.
Now, let's draw the velocity-time graph based on the analysis above.
|
340 ______________|_____________
| / \
| / \
| / \
240 |______________________________
| 30 50 110 130
Note: The x-axis represents time in seconds, and the y-axis represents velocity in meters per second.
(b) Using the graph, we can calculate the required values:
(i) Maximum speed attained during the motion:
The maximum speed is where the graph reaches its highest point. From the graph, the maximum speed is 340 m/s.
(ii) Average retardation as the body is being brought to rest:
The average retardation can be calculated using the formula a = Δv / t, where Δv is the change in velocity and t is the time taken.
Here, Δv = 340 m/s - 0 m/s = 340 m/s, t = 20 seconds.
So, the average retardation is a = 340 m/s / 20 s = 17 m/s^2.
(iii) Total distance traveled during the first 50 seconds:
To find the total distance traveled during the first 50 seconds, we need to calculate the area under the velocity-time graph for that time interval.
From the graph, the shape under the curve can be divided into a rectangle and a triangular segment.
Rectangle area = base * height = 20 s * 340 m/s = 6,800 m.
Triangle area = (1/2) * base * height = (1/2) * 30 s * 240 m/s = 3,600 m.
Total distance = rectangle area + triangle area = 6,800 m + 3,600 m = 10,400 m.
(iv) Average speed during the same interval as in (iii):
Average speed is calculated by dividing the total distance traveled by the total time taken.
Total distance = 10,400 m, Total time = 50 seconds.
Average speed = Total distance / Total time = 10,400 m / 50 s = 208 m/s.