A car runs at a constant velocity of 15m/s for 300sec and then accelerates uniformly to a velocity of 25m/s over a period of 20s. This velocity is mantained for 300s before the car is brought to rest with uniform deceleratiom in 30s. Draw a velocity time graph to represent the journey described above.
IMG_20220210_072404_2.jpg
a) the acceleration while the velocity changes from 15m/s to 25m/s
Solution
a=∆v/∆t=V2-V1/t2-t1
=25-15/320-300=10/20
1/2m/s or 0.5m/s
b) total distanced travelled
Solution
L×B + ½(a+b)h + L×B + ½b×h
300×15 + ½(15+25)20 + 300×25 + ½×30×25
4500 + 400 + 7500 + 375
=12775m
C) Average speed = total distanced cover/time taking
=12775/650
19.6m/s
I need an answer
To draw a velocity-time graph, we need to identify the key points during the journey and plot them on the graph.
Step 1: The car runs at a constant velocity of 15 m/s for 300 seconds. This means the velocity remains constant throughout this time period. We can represent this as a horizontal line on the graph at 15 m/s.
Step 2: The car then accelerates uniformly over a period of 20 seconds until it reaches a velocity of 25 m/s. To represent this, we can draw a straight line with a positive slope starting from the end point of the previous line (15 m/s at 300 seconds) and ending at (25 m/s, 320 seconds).
Step 3: The car maintains a velocity of 25 m/s for 300 seconds, so we can draw a horizontal line at this velocity from (25 m/s, 320 seconds) to (25 m/s, 620 seconds).
Step 4: Finally, the car is brought to rest with uniform deceleration in 30 seconds. We can represent this as a straight line with a negative slope starting from (25 m/s, 620 seconds) and ending at the point where the velocity reaches zero.
Here is how the velocity-time graph would look:
^
|
30 - +
|
+------------------------------------------------------------------
|
|
25 - +----------------------------+----------------------------------
| |
| |
15 - +----------------------------+----------------------------------
| |
0 300 620 950 --> Time (seconds)
Note: The exact time values for the points may not match the scale provided, but the overall shape of the graph should resemble the one shown above.
To draw a velocity-time graph representing the journey described, we will break it down into three phases:
1. Constant Velocity Phase:
The car runs at a constant velocity of 15 m/s for 300 seconds. Since the velocity is constant, the graph will be a straight horizontal line at a height of 15 m/s for the duration of 300 seconds.
Velocity (m/s)
^
15 -----|--------------------- 300s
2. Uniform Acceleration Phase:
The car then accelerates uniformly from a velocity of 15 m/s to 25 m/s over a period of 20 seconds. To represent this, we will draw a straight line with positive slope from the end point of the constant velocity phase to the new velocity value.
Velocity (m/s)
^
15 -----|\
| \
| \ 20s
| \
25 ----|-----\-------------300s
3. Uniform Deceleration Phase:
After maintaining the velocity of 25 m/s for 300 seconds, the car is brought to rest with uniform deceleration in 30 seconds. To represent this phase, we will draw a straight line with a negative slope from the end point of the uniform acceleration phase to the x-axis (zero velocity).
Velocity (m/s)
^
15 -----|\
| \
| \
| \ 300s
25 ----|-----\--------\--------\-- 30s
Time (s)
In this graph, the x-axis represents time, and the y-axis represents velocity. The graph shows the change in velocity of the car over time during its journey.