Question

Train starts from rest with this from station with the acceleration of 0.2 metre per second square on the stage straight track and then comes to rest after attending maximum speed on the another station due to retardation of 0.4 metre per second square

Answer 1

Answer:

Distance will be 216 km.

Explanation:

Since, from the question we will get that the train accelerates with acceleration of a = 0.2 m/s^2, and if we let t = t1 and v= v1.

So, applying newtons laws of motion v=u + at, since we know that the v1=0.2t1 or t1= 0.2 v1 /2 (1)

.

Again, applying v2=v1+a2t2(when train is retarding and s =s1).

S=ut + 1/2at^2( where u will be zero).

s1=1/2 x 0.2 x (t1)^2

.

So, now v2=v1+a2t2.

0 = v1+0.4t2

t2=–v1/0.4 (2)

Now, applying them in the equation s2=v1t2 +(1 /2) x a2(t2)^2.

So, the total time will be t=t1+t2 = 30 min = 1800 seconds.

So, we get that t1=1200 secs and t2 = 600 secs.

So, then substituting the values of a, v and t in (1) and (3),

s1 = 144 km and s2= 72 km

Total distance will be s= s2+ s2.

s=216km

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