A car runs at a constant speed of 15m/s for 300s and then accelerates uniformly to a speed of 25m/s over a period of 20s this speed is maintained for 300s before the car is brought to rest with uniform deceleration in 30s
Sketch the velocity time graph to represent the journey described above
since v = at, the graph will consist of
a horizontal line segment
a sloping line segment with slope=25
another horizontal bit
a sloping bit with slope = -(2/3)*25
NO IDEA
PLEASE ANSWER IT FOR PLS
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Well, when it comes to graphing, I'm more of a stand-up comedian than a sketch artist, but I'll give it a shot!
First, we have the car running at a constant speed of 15m/s for 300s. So we'll draw a horizontal line at 15m/s for the first 300s.
Next, the car accelerates uniformly to a speed of 25m/s over a period of 20s. That means we'll have an increasing slope from 15m/s to 25m/s over those 20s.
Once the car reaches 25m/s, it maintains that speed for 300s. So we'll draw a horizontal line at 25m/s for those 300s.
Finally, the car is brought to rest with uniform deceleration in 30s. This means we'll have a decreasing slope from 25m/s to 0m/s over those 30s.
So to sum it up, you'll see a flat line at 15m/s for 300s, then an increasing slope from 15m/s to 25m/s over 20s, followed by a flat line at 25m/s for 300s, and finally a decreasing slope from 25m/s to 0m/s over 30s.
Remember, I'm no Picasso, but I hope this gives you a good laugh and helps you visualize the journey!
To sketch the velocity-time graph for the car's journey, we need to break down the journey into different intervals and determine the corresponding velocity values for each interval.
1. Constant Speed: 15 m/s for 300 seconds
During this interval, the velocity remains constant at 15 m/s. So, we represent this as a horizontal line on the graph at the height of 15 m/s for the duration of 300 seconds.
2. Acceleration: Uniform acceleration from 15 m/s to 25 m/s over 20 seconds
Here, the car is accelerating uniformly from 15 m/s to 25 m/s. We can use the equation of uniform acceleration:
v = u + at
where:
v = final velocity (25 m/s)
u = initial velocity (15 m/s)
a = acceleration (unknown)
t = time taken (20 seconds)
We can rearrange the equation to solve for acceleration:
a = (v - u) / t
a = (25 - 15) / 20
a = 10 / 20
a = 0.5 m/s^2
Now, we can calculate the displacement during acceleration using the formula:
s = ut + 0.5at^2
where:
s = displacement
u = initial velocity (15 m/s)
t = time taken (20 seconds)
a = acceleration (0.5 m/s^2)
s = (15 * 20) + 0.5 * 0.5 * (20^2)
s = 300 + 0.5 * 0.5 * 400
s = 300 + 0.5 * 0.5 * 400
s = 300 + 0.5 * 200
s = 300 + 100
s = 400 meters
During this interval, the velocity gradually increases from 15 m/s to 25 m/s while covering a displacement of 400 meters. So, we represent this as a straight line with a positive slope on the graph.
3. Constant Speed: 25 m/s for 300 seconds
The car maintains a constant speed of 25 m/s during this interval. So, we represent this as a horizontal line on the graph at the height of 25 m/s for the duration of 300 seconds.
4. Deceleration: Uniform deceleration from 25 m/s to 0 m/s over 30 seconds
Here, the car is decelerating uniformly from 25 m/s to 0 m/s. We can use a similar equation of uniform acceleration to calculate the deceleration:
v = u + at
where:
v = final velocity (0 m/s)
u = initial velocity (25 m/s)
a = deceleration (unknown)
t = time taken (30 seconds)
We can rearrange the equation to solve for deceleration:
a = (v - u) / t
a = (0 - 25) / 30
a = -25 / 30
a = -0.8333 m/s^2
During this interval, the velocity gradually decreases from 25 m/s to 0 m/s. So, we represent this as a straight line with a negative slope on the graph.
By combining all these intervals on the graph, you will have a velocity-time graph that represents the journey described.