A train starting from rest accelerates uniformly for 100 seconds, runs at a constant speed for 5 minutes and then comes to a stop with uniform retardation.During this motion it covers a distance of 4.25kms.Find its constant speed, acceleration, and retardation

Q) A train starting from rest accelerates uniformly for 100 seconds, runs at a constant speed for 5 minutes and then comes to a stop with uniform retardation.During this motion it covers a distance of 4.25kms.Find its constant speed, acceleration, and retardation

 

Sol) Here there are three cases

 

a) for the first case when acceleration is uniform

 

s₁ = ut + 1/2 at²

 

here t = 100s, u =0

 

so s₁ = 0.5 a *100*100 = 5000a

 

b) For second part of the journey when speed is uniform

 

s₂ = vt = v*5*60 = 300v (as t = 5min = 300s)

 

c) for final part of journey (uinform retardation)

 

s₃ = ut - 1/2 a t² = s₁

 

Now s₁ + s₂ + s₃ = 4.25 km = 4250 m

 

or 5000a + 300v + 5000a = 4250

 

so 10000a + 300v = 4250   ...(1)

 

Now in he first case using v-u/t = a

 

or a = v/100      ...(2)

 

Thus, using (2)n (1) we obtain

 

100v + 300v = 4250

 

or v = 4250/400 =  10.625m/s

 

and a = v/100 = 10.625/100 = 0.10625 m/s²