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A train starts from rest and moves with a constant acceleration for the first 1 km. For the next 3 km, it has a constant velocity and for the last 2 km, it moves with constant retardation to come to rest after a total time of motion of 10 min. Find the maximum velocity and the three time intervals in the three types of motion.
A train starts from rest and moves with a constant acceleration for the first 1 km.  For the next 3 km,  it has a constant velocity and for the last 2 km,  it moves with constant retardation to come to rest after a total time of motion of 10 min.  Find the maximum velocity and the three time intervals in the three types of motion.


4 years ago

Shaswata Biswas
132 Points
							Suppose the velocity of the train increases from 0 to v in time t1, then moves with constant speed v upto time t2 and finally its velocity decreases from v to 0 between time interval t2 and t.For motion between t=0 and t=t1 let acceleration be \alpha.Then, \alpha = (v-0)/(t1-0)  = v/t1=> t1 = v/\alphaSimilarly, for retardation \beta\beta = (v-0)/(t-t2)  = v/(t-t2)=> t-t2 = v/\beta=> t1+t-t2 = v/\beta + v/\alpha=> t2-t1 = t - v{1/\alpha + 1/\beta}Draw a graph of the motion of the train, from where we can find the displacement of the train.Now, for total displacement S = 1+3+2 = 6 km, and total time taken t=10 min = 1/6 h, we have.S = v.t1/2 + v(t2-t1) + v(t-t2)/2S = v2/2\alpha + v[t- v{1/\alpha + 1/\beta}] + v2/2\betaS/v = v{1/\alpha + 1/\beta}/2 + t - v{1/\alpha + 1/\beta} = t - v{1/\alpha + 1/\beta}/2or, t = S/v + v{1/\alpha + 1/\beta}/2Puutting the values of the variables, the max. velocity attained "v" can be claculated.THANKS

4 years ago
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• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions