Question

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

Answer 1

Answer: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/\alpha

Similarly, 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)/2

S = v2/2\alpha + v[t- v{1/\alpha + 1/\beta}] + v2/2\beta

S/v = v{1/\alpha + 1/\beta}/2 + t - v{1/\alpha + 1/\beta} = t - v{1/\alpha + 1/\beta}/2

or, t = S/v + v{1/\alpha + 1/\beta}/2

Puutting the values of the variables, the max. velocity attained "v" can be claculated.

Explanation:hope this helps u plz mark me as the brainliest