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.

are there answer choices to this question? this seems like a question where you have options to choose from.

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg