Sir, how to find the of trapezium under velocity-time graph. Supposed, A train starts from station P and accelerates uniformly for 2min reaching a speed 48kmp/hr. It continues at this speed for 5min and then is retarded uniformly for a further 3min to come to rest at station Q. Find (1)the distance PQ in km; (2) the average speed of the train; (3)the acceleration in m/sec square; (4) the time taken to cove half the distance between P and Q
I'll describe the graph that I am picturing. The x axis is time and the y axis is velocity. (I trust you didn't mean to say 48 kmp/hr. The p would mean per and the / would also mean per.) Since the acceleration is uniform, the line for the speeding up and slowing down sections will be straight. The first section of the velocity-time graph is a straight line starting at 0,0 and ending at 2min,48km/hr. The next section is a horizontal line that continues from 2min,48km/hr to 7min,48km/hr. The third section goes from the last point to 10min,0.
1. PQ is the sum of the distances traveled in the 3 sections. So it is the area under the lines of all 3 sections. The middle one is easiest. The velocity is constant at 48 km/min and the time is 5 min.
D2 = velocity*time = (48km/hr)*5min*1hr/60min = 4km
In the first section, the time times the arithmetic average of the starting speed and ending speed would yield the distance traveled during the section. This only works because the acceleration is uniform.
D1 = average velocity*time = ((0+48km/hr)/2)*2min*1hr/60min = 0.8 km
The same approach works for the 3rd section. I will leave working that out for you to do. And then you just add D1+D2+D3.
2. So we need the average speed, Vave, of this train (#1) during the trip from P to Q. Imagine a second train going cross country that is not going to stop at P or Q. It will go at constant speed the whole time. If 10 minutes passes between this train passing P and passing Q, the speed of train #2 equals the average speed, Vave of train #1. That's what average speed means: a constant speed that would take you from point A to point B in the same time that was actually required with the variations in speed that existed in the real case.
So what constant speed would carry train 2 past these 2 points in 10 minutes?
V2 = distance/time = your answer to Question1 / (10min*1hr/60min) = ____ km/hr
And the average speed of train 1 = the above V2
3. Since the question asks for the answer with units m/s^2, we should convert the speed in km/hr to m/s and time in minutes to s.
48km/hr*(1000m/1km)*(3600s/1hr) = speed in m/s You can do the math.
2min*(60s/1min) = time in seconds
The acceleration during the first section is given by the kinematic formula
Vf = Vi + a*t
Plug in the final and initial velocities, with units of m/s, and the time in seconds; and solve for a.
4. You now know the total distance and the distance traveled in the first section. Find 1/2 the total distance and subtract the distance traveled in the first section. The result is how much farther it has to go during the second section to reach the halfway point. Call that distance Dhalf. The velocity is constant in section 2, so
48km/min = Dhalf/time
Solve for time.
I hope this helps,