Atrain starts from rest and travels for 5.0 s with a uniform acceleration of 1.5 m/s 2. how far does the train travel in this time interval?

Given: Initial velocity of train (u) = 0

Acceleration of train (a) = 1.5 m/s²

Travel time (t) = 5.0 s

To Find: The distance traveled by train in the given time interval

Formula Used: 2nd kinematic equation of motion

s = ut + (1/2) at²

Here, all alphabets are in their usual meanings.

Now, using equation (i), we shall calculate the value of s

                     s = 0×5.0 + (1/2) ×1.5×(5.0)² m

or,                 s = 18.75 m

Hence, the required traveled distance by the train will be 18.75 m.

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.