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?

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

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2