Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
A train starts from rest and moves with a constant acceleration of 2m/s^2 for half a minute. The brakes are then applied and the train comes to rest in one minute after applying brakes. Find (a) the total distance moved by the train, (b) the maximum speed attained by the train and (c) the position of the train at the half the maximum speed.
Let maximum speed attained by train is v, distance covered is s while accelerating and S while deaccelerating. v=0+2(30)=60m/ss=0+(1/2)(2)(900)s=900m0=60-a(60)a=1m/s²S=60(60)-½(60)²1S=3600-1800S=1800m
Sorry I misunderstood the c part so here is the answer to (c) partDistance covered at half the maximum speed30²=0+2(2)s1s1=900/4=225m
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -