Question

A freight train traveling with a velocity of −4.0 m/s begins backing into a train yard. If the train's average acceleration is −0.27 m/s^2 , what is the train's velocity after 17 s?

Answer 1

Answer:

−8.59 m/s.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = −4.0 m/s

Acceleration (a) = −0.27 m/s²

Time (t) = 17 s

Final velocity (v) =?

Acceleration is defined as the change of velocity with time. Mathematically, it is expressed as:

Acceleration = change of velocity / time

Acceleration = (final velocity – Initial velocity) / time

a = (v – u) / t

With the above formula, we can determine the velocity of the train after 17 s. This can be obtained as follow:

Initial velocity (u) = −4.0 m/s

Acceleration (a) = −0.27 m/s²

Time (t) = 17 s

Final velocity (v) =?

a = (v – u) / t

−0.27 = (v – –4 ) / 17

−0.27 = (v + 4 ) / 17

Cross multiply

−0.27 × 17 = v + 4

−4.59 = v + 4

Collect like terms

−4.59 – 4 = v

−8.59 = v

v = −8.59 m/s

Thus, the velocity of the train after 17 s is −8.59 m/s.