Amonorail train starts from rest on a curve of radius 400 m and

accelerates at the constant rate a_t. if the maximum total acceleration

of the train must not exceed 1.5 m/s^2, determine (a) the shortest

distance in which the train can reach a speed of 72 km/h, (b) the

corresponding constant rate of acceleration a_t.

(a) Shortest distance traveled by train = 178.89 m

(b) Constant acceleration will be      

Explanation:

We have given speed v = 72 km /hr =

Radius of path = 400 m

Total acceleration

We know that normal acceleration is given by

(b) Total acceleration is given by

So corresponding acceleration will be

(a) As the train starts from rest its initial velocity u = 0 m/sec

Final velocity v = 20 m/sec

Acceleration

From third equation of motion we know that

s = 178.89 m

answer:

d. attempt to find the owner, then leave your information if you cannot find them.

explanation:

d. represents both (b) and (c).

more logical as well.

b and c are to complicated.

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