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 small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

-reverse card

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form