Freight trains can produce only relatively small acceleration and decelerations. (a) What is the final velocity (in m/s) of a freight train that accelerates at a rate of 0.0400 m/s2 for 8.00 min, starting with an initial velocity of 3.00 m/s? 22.2 Correct: Your answer is correct. m/s (b) If the train can slow down at a rate of 0.550 m/s2, how long (in s) will it take to come to a stop from this velocity? Incorrect: Your answer is incorrect. s (c) How far (in m) will it travel in each case?

Question

Physics

Zachary Maynard

3 April, 10:46

Freight trains can produce only relatively small acceleration and decelerations. (a) What is the final velocity (in m/s) of a freight train that accelerates at a rate of 0.0400 m/s2 for 8.00 min, starting with an initial velocity of 3.00 m/s? 22.2 Correct: Your answer is correct. m/s (b) If the train can slow down at a rate of 0.550 m/s2, how long (in s) will it take to come to a stop from this velocity? Incorrect: Your answer is incorrect. s (c) How far (in m) will it travel in each case?

+1

Answers (1)

Know the Answer?

Micah Stephenson · Answer 1

Answer

Initial velocity of the freight train = V₁ = 3 m/s

Acceleration of the freight train = a₁ = 0.04 m/s²

Time period the freight train accelerates for = T₁

= 8 min = 480 sec

Velocity of the freight train after accelerating for 8 min = V₂

V₂ = V₁ + a₁T₁

V₂ = 3 + (0.04) (480)

V₂ = 22.2 m/s

Distance traveled by the freight train while accelerating = D₁

D₁ = V₁T₁ + 0.5 * a₁T₁²

D₁ = (3) (480) + 0.5 * (0.04) (480) ²

D₁ = 6048

Deceleration of the freight train = a₂ = - 0.55 m/s²

Final velocity of the freight train = V₃ = 0 m/s (Comes to a stop)

Time taken by the freight train to come to a stop = T₂

V₃ = V₂ + a₂T₂

0 = 22.2 + (-0.55) T₂

T₂ = 40.36 sec

Distance traveled by the freight train during deceleration = D₂

D₂ = V₂T₂ + 0.5a₂T₂²

D₂ = (22.2) (40.36) + 0.5 (-0.55) (40.36) ²

D₂ = 448.04 m