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Grade: 9
        
A train starts from rest from station with acceleration 0.2 metre per second square on a straight track and then comes to rest after attaining maximum speed on another station due to retardation of 0.4 metre per second square if total time is half hour then distance between two stations is?
 
Ans(216km) please explain clearly
4 months ago

Answers : (2)

Arun
22984 Points
							
If A is the acceleration of the train and B is the retardation of the train, then the total distance travelled is given by
 [(AB) / (A+B)]. (T^2/2) 
Here A = 0.2m/s^2 and B = 0.4m/s^2 and 
T = 1/2 hr = 1800 sec 
Therefore x= [(0.2) (0.4)/0.2+0.4)]x1800x1800/2 
= 216000m 
= 216km 

 
 
4 months ago
Khimraj
3008 Points
							
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
3 months ago
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