A bus moves from rest with a uniform acceleration of 2ms for d first 10s. If den accelerates at a uniform rate of 1ms for another 15s . it continues at a constant speed for 70s and finally comes to rest in 20s by uniform acceleration.
(A) draw d velocity time graph of d motion using d information given above. From ur graph otherwise calculate d total distance traveled
draw the graph. The area underneath the graph is distance.
To draw the velocity-time graph of the motion described, we will break the motion into different time intervals and calculate the velocity at the end of each interval.
First 10 seconds:
The bus starts from rest, so the initial velocity is 0 m/s.
The acceleration is given as 2 m/s².
Using the equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can calculate the velocity at the end of this interval:
v = 0 + 2 * 10 = 20 m/s.
Next 15 seconds:
The bus continues to accelerate at a uniform rate of 1 m/s².
The initial velocity for this interval is the final velocity of the previous interval, which is 20 m/s.
Using the same equation of motion, we can calculate the velocity at the end of this interval:
v = 20 + 1 * 15 = 35 m/s.
The bus then continues at a constant speed for 70 seconds, so the velocity remains constant at 35 m/s.
The last 20 seconds:
The bus comes to rest, so the final velocity is 0 m/s.
The acceleration is given, but we need to determine the initial velocity for this interval.
Since the velocity remains constant at 35 m/s before decelerating, the initial velocity for this interval is also 35 m/s.
Using the equation of motion, v = u + at, we can calculate the acceleration during this interval:
0 = 35 + a * 20
a = -35/20 = -1.75 m/s² (negative sign indicates deceleration).
Now we can plot the velocity-time graph using this information:
Time (s) Velocity (m/s)
0 0
10 20
25 35
95 35
115 0
The total distance traveled can be calculated by finding the area under the velocity-time graph. Since the graph consists of multiple sections, we need to calculate the area for each section separately and sum them up.
Area for the first section (rectangle):
Length = 10 s
Width (velocity) = 20 m/s
Area = length * width = 10 * 20 = 200 m
Area for the second section (rectangle):
Length = 15 s
Width (velocity) = 35 m/s
Area = length * width = 15 * 35 = 525 m
Area for the third section (rectangle):
Length = 70 s
Width (velocity) = 35 m/s
Area = length * width = 70 * 35 = 2450 m
Area for the fourth section (triangle):
Base = 20 s
Height (average velocity) = (35 + 0)/2 = 17.5 m/s
Area = (1/2) * base * height = (1/2) * 20 * 17.5 = 175 m
Total distance traveled = 200 + 525 + 2450 + 175 = 3350 m.